6 31-42 SDOF Free and Forced Vibration: Duffing Equation ... · length of the pendulum (= 1 m) ,...

NPTEL – Mechanical Engineering – Nonlinear Vibration

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Module 6

APPLICATIONS

NONLINEAR VIBRATION OF MECHANICAL SYSTEMS

6 Applications

31-42 SDOF Free and Forced Vibration: Duffing Equation, van der pol’s Equation: Simple or primary resonance, sub-super harmonic resonance. Parametrically excited system- Mathieu-Hill’s equation, Floquet Theory, Instability regions; Multi-DOF nonlinear systems and Continuous system, System with internal resonances

6 A Free

Vibration of nonlinear Systems

1 Single degree of freedom Nonlinear conservative systems with Cubic nonlinearities.

2 Single degree of freedom nonlinear conservative systems with quadratic and Cubic and nonlinearities.

3 Single degree of freedom non-conservative systems: viscous damping, quadratic and Coulomb damping

4 non-conservative systems: Negative damping, van der Pol oscillator, simple pendulum with quadratic damping

6 B Forced

nonlinear Vibration

1 Single degree of freedom Nonlinear systems with Cubic nonlinearities: Primary Resonance

2 Single degree of freedom nonlinear systems with Cubic nonlinearities: Nonresonant Hard excitation

3 Single degree of freedom Nonlinear systems with Cubic and quadratic nonlinearities and self sustained oscillations

4 Multi-degree of freedom nonlinear systems

6 C nonlinear Vibration

of Parametrically

excited system

1 Parametrically excited system: Floquet theory, Hill’s infinite determinant

2 Parametric Instability region: sandwich beam vibration

3 Base excited magneto-elastic cantilever beam with tip mass 4 System with internal resonance: Two-mode interaction: Base excited

cantilever beam with tip mass at arbitrary position



Module 6 Lect 1

Free Vibration of Nonlinear Conservative system

In this lecture we will learn about the free vibration response of nonlinear conservative systems. Initially the qualitative analysis will be demonstrated and later by using one of the perturbation methods the free vibration response of the single degree of freedom system will be illustrated using different examples.

Qualitative Analysis of Nonlinear Systems

Consider the nonlinear conservative system given by the equation

( ) 0u f u+ = (6.1.1)

Multiplying u in Eq. (6.1.1) and integrating the resulting equation one can write as

( ) ( )( ( ))

Or, ( )

Or, ( )

uu uf u dt h

d u d uu dt f u dt h

dt dtudu f u du h

+ =

+ =

+ =

∫

∫ ∫∫ ∫

(6.1.2)

Or., 1 2 ( ) , where, ( ) ( )2

u F u h F u f u du+ = = ∫ (6.1.3)

This represents that the sum of the kinetic energy and potential energy of the system is constant. Hence, for particular energy level h, the system will be under oscillation, if the potential energy ( )F u is less than the total energy h. From the above equation, one may plot the phase portrait or the trajectories for different energy level and study qualitatively about the response of the system using the following equation.

( )2 ( )u h F u= − (6.1.4)

It may be noted that velocity exists, or the body will move only when ( )h F u> . One will obtain equilibrium points corresponding to ( )h F u= or when ( ) ( ) 0F u f u′ = = . For minimum potential energy a center will be obtained and for maximum potential energy a saddle point will be obtained. The trajectory joining the two saddle points is known as homoclinic orbit. The response is periodic near the center.

Example 6.1.1 Perform qualitative analysis of spring-mass system with a soft spring. Take mass of the system as 1unit, linear stiffness 1 unit and stiffness corresponding to cubic nonlinear term as 0.1 unit.



Solution: In this case the equation of motion of the system can be given by

30.1 0x x x+ − = (6.1.5)

For this system 3 2 41 1( ) ( ) ( 0.1 )2 40

F x f x dx x x dx x x= = − = −∫ ∫ (6.1.6)

Figure 6.1.1 shows the variation of potential energy F(x) with x. It has its optimum values corresponding to ( )( ) ( ) 0F x f x′ = = 0 or 10x = ± . While x equal to zero represents the system with

minimum potential energy, the other two points represent the equilibrium points with maximum potential energy. Now by taking different energy level h , one may find the relation between the

velocity v and displacement x as 2 42( ( )) 2( (0.5 0.025 ))v x h F x h x x= = − = − −

Fig. 6.1.1: Potential well (curve with blue colour) and phase portrait (red colour) showing saddle point (S) and center (C) corresponding to maximum and minimum potential energy.

Now by plotting the phase portrait one may find the trajectory which clearly depicts that the equilibrium point corresponding to maximum potential energy is a saddle point (marked by point S) where the equilibrium point is unstable and the equilibrium point corresponding to the minimum potential energy is stable center type (marked by point C). Clearly the orbit joining the points S and S is homoclinic orbit. Depending on different initial conditions i.e the total energy of the system, near the center one will obtain periodic orbits.

Example 6.1.2: Perform qualitative analysis for a simple pendulum.

Solution: The governing equation of motion of the system is given by (Eq. 2.1.7)

S C S



sin 0 or sin 0gml mgl

θ θ θ θ+ = + = (6.1.7)

So, ( ) ( ) sin cosg gF f d dl l

= = = −∫ ∫θ θ θ θ θ θ (6.1.8)

Hence, the potential function ( )F θ has a minima corresponding to θ equal to zero and maxima corresponding to θ equal to odd multiple of π . So, the equilibrium point near 0=θ is a center and near

180= θ is a saddle point. The phase portrait for different energy level h is given in Figure 6.1.3 which is obtained by using the following relation.

2( ( )) 2 cosgv h F hl

= = − = +

θ θ θ (6.1.9)

Figure 6.1.3: Phase portrait for the motion of a simple pendulum Approximate solution method

C C C C C

S S S S

Fig. 6.1.2: Simple pendulum



Let us consider the Duffing equation with cubic nonlinear term for the free vibration study of a nonlinear system.

22 302 0 d u u u

dtω ε+ + = (6.1.10)

Using method of multiple scales, the solution of this equation can be given by

2 30 1 2 ( ) u u u u oε ε ε= + + + (6.1.11)

Taking the time scale ε= n

nT t (6.1.12) 2

0 1 2 ................, nn

d D D D Ddt T

ε ε ∂= + + + =

∂ (6.1.13)

( )2

2 2 20 0 1 1 0 22 2 2 ................d D D D D D D

dtε ε= + + + + (6.1.14)

Substituting Eqs. (6.1.11-6.1.14) in Eq. (6.1.10) and separating terms with different order of ε one obtains the following equations. Order of 0ε

2 20 0 0 0 0D u uω+ = (6.1.15)

Order of 1ε

2 2 30 1 0 1 0 1 0 02D u u D D u uω+ = − − (6.1.16)

Order of 2ε 2 2 2 20 2 0 2 0 1 1 0 2 0 1 0 0 12 2 3D u u D D u D D u D u u uω+ = − − − − (6.1.17)

Solution of Eq. (6.1.15) can be given by

0 0 0 00 1 2 1 2( , , ) ( , , )i T i Tu A T T e A T T eω ω−= + (6.1.18)

Substituting Eq.(6.1.18) in Eq. (6.1.16) one obtains

0 0 0 020

32 2 310 1 0 1 2 3[ ] ω ωωω+ = − − ++ i T i Ti D A AD e AAu u e cc (6.1.19)

To eliminate the secular term (coefficient of 0 0ωi Te ) term marked in pink colour should be equated to zero. Hence, 2

0 12 3 0i D A A Aω + = (6.1.20) From Eq. (6.1.19), the solution of 1u can be written as



0 0

33

1 208

i TAu e ccω

ω= + (6.1.21)

Now Substituting 1 exp( )2

A a iβ= (where a and β are real number) in Eq. (6.1.20) and separating the

real and imaginary parts one obtains

1

0,aT∂

=∂

and 30

1

3 08

a aTβω ∂

− + =∂

(6.1.22)

Hence, a is not a function of 0T and 1T . So up on integration Eq. (6.1.22) can be written as

2( ),a a T= and 21 0 22

0

3 ( )8

a T Tβ βω

= + (6.1.23)

Substituting Eq. (6.1.18) and Eq. (6.1.21) in Eq. (6.1.17), one can write

0 0 0 0 0 0

Secular t

3 52 2 4 50 2 0 2 2

e

3 2

0 2 20 00

m

2

r

21 38

128 8

5 i T i T i TD A Aiu u e A Ae A e cD A cω ω ωωω ω

ωω

+ = − + − +

−

(6.1.24)

For the secular term to be zero one can write 3 2

0 2 20

1528A Ai D Aωω

− (6.1.25)

The solution of remaining part of Eq. (6.1.24) can be written as

0 0 0 0

55 34

2 4 40 0

2164 64

i T i TAu e A Ae ccω ω

ω ω= − + + (6.1.26)

Now Substituting 1 exp( )2

A a iβ= (where a and β are real number) in Eq. (6.1.25) and separating the

real and imaginary parts one obtains

2

0,aT∂

=∂

and 40 2

2 0

15256

aTβω

ω∂

− =∂

(6.1.27)

So, a is a constant. Now using Eqs. (6.1.23, 6.1.27) one can write 4

0 230

15256

a Tβ γω

= − + (6.1.28)

Hence, from Eq. (6.1.23)

2 41 23

0 0

3 158 256

a T a Tβ γω ω

= − + (6.1.29)

Hence the solution of the system can be written as 23 2 5

322 400 0

211cos( ) cos3( ) cos5( ) ( )3232 1024

aa au a t t t oε εεω γ ω γ ω γ εωω ω

−= + + + + + +

(6.1.30)



where, 2 4

2 30 3

0 0

3 15 ( )8 256

ω ω ε ε εω ω

= + − +a a o (6.1.31)

From Eq. (6.1.31) it may be noted that the frequency of oscillation is a function of amplitude oscillation. It may be recalled that in case of linear system frequency is independent of amplitude of oscillation. For, the simple pendulum taking 0 /g Lω = , where g is the acceleration due to gravity and L is the length of the pendulum (= 1 m) , the variation of frequency with amplitude is shown in Fig. 6.1.4. The phase portrait for the periodic response considering only the first order solution is shown in Fig. 6.1.5.

Exercise Problems: 1. Perform qualitative analysis for the following nonlinear systems. Using equation (6.1.30) plot the response of the system and compare both the results. (a) µ +4µ+𝜇3 =0 (c) µ +9µ+0.5 𝜇2+ 0.1𝜇3 =0

(b) µ +µ+𝜇2 =0 (d) µ +100µ+ 10𝜇3 =0

(e) µ +µ+0.1 𝜇3+ 0.05𝜇3 =0

Fig. 6.1.4: Frequency amplitude relation for a simple pendulum

Fig. 6.1.5: Phase portrait showing the periodic motion for a simple pendulum



Module 6 Lecture 2

Free vibration of nonlinear single degree of freedom conservative systems with quadratic and cubic nonlinearities. In this lecture the free vibration response of a nonlinear single degree of freedom system with quadratic and cubic nonlinearities will be discussed with numerical examples. As studied in module 3, the equation of motion of a nonlinear single degree of freedom system with quadratic and cubic nonlinearities can be given by

2 2 30 2 3 0x x x xω εα εα+ + + = (6.2.1)

Here 0ω is the natural frequency of the system 2α and 3α are the coefficient of the quadratic and cubic nonlinear terms. Also ε is the book-keeping parameter which is less than 1. Using method of multiple scales the solution of this equation can be written as

( ) ( ) ( ) ( ) .....,......,,,......,,,......,,; 21033

21022

2101 +++= TTTxTTTxTTTxtx εεεε (6.2.2)

Using different time scales ,, 10 TT and 2T where n

nT tε= and Eq. (6.2.2) in Eq. (6.2.1) and separating the terms with different order of ε one can write the following equations.

Order of 1ε

2 20 1 0 1 0D x xω+ = (6.2.3)

Order of 2ε

2 2 20 2 0 2 0 1 1 2 12D x x D D x xω α+ = − − (6.2.4)

Order of 3ε

2 2 2 30 3 0 3 0 1 2 1 1 0 2 1 2 1 2 3 12 2 2D x x D D x D x D D x x x xω α α+ = − − − − − (6.2.5)

The solution of (6.2.3) can be written as

1 1 2 0 0 0 0( , ) exp( ) exp( )x A T T i T A i Tω ω= + − . (6.2.6)



Here A is an unknown complex function and A is the complex conjugate of A . Substituting (6.2.3) into (6.2.4) leads to

Secular term

2 2 20 2 0 2 2 0 00 1 0 0 exp(2( ) )2 expD x x A i T AA cci D A i Tω α ωω ω + = − − + +

(6.2.7)

Here cc denotes the complex conjugate of the preceding terms. To have a bounded solution one should eliminate the secular term and hence

11

0dAD AdT

= = (6.2.8)

Therefore A must be independent of 1T . With 1 0D A = the particular solution of (6.2.7) can be written as

2

2 22 0 02 2

0 0

exp(2 )3

Ax i T AA ccα αωω ω

= − + (6.2.9)

Substituting the expression for 1x and 2x from equation (6.2.6) and (6.2.9) into (6.2.5) and recalling

that 1 0D A = we obtain

( )2 2

22 3 00 2 0 02

0

2 20 3 0 3

2 233 0 2

0 020

Secular Te3 2 exp

10 92 exp3

rm

(3 )3

i D A A A iD x x

A i

T

T cc

αω

α ω α

α ωω ω

ω

ω

ω

+ = −

+− +

−−

(6.2.10)

To eliminate the secular terms from 3x , we must put

2 2

23 0 20 2 2

0

9 102 03

i D A A Aα ω αωω−

+ = (6.2.11)

Using 1 exp( )2

A a iβ= where a and β are real function of 2T in Eq. (6.2.11) and separating the result

into real and imaginary parts, one obtains

0aω ′ = and 2 2

32 3 00 2

0

10 9 024

a aα α ωω βω−′+ = (6.2.12)



where the prime denotes the derivative with respect to 2T . As 0a′ = , a is a constant and 2 2

32 3 020 0

10 924

aa

α α ωβω ω−′ = − or

2 223 0 2

2 030

9 1024

a Tα ω αβ βω−

= + (6.2.13)

Here 0β is a constant. Now using 22T tε= one may write

2 2

2 23 0 203

0

9 101 exp2 24

A a i a t iα ω α ε βω

−= +

(6.2.14)

Substituting Eq. (6.2.14) in the expressions for 1x and 2x in Eqs. (6.2.6), (6.2.9) and (6.2.2), one obtains

2 232

0 020

1cos( ) 1 cos(2 2 ) ( )2 3ax a t t Oε αε ω β ω β εω

= + − − + + (6.2.15)

Here 2 2

2 2 33 0 20 4

0

9 101 ( )24

a Oα ω αω ω ε εω

−= + +

(6.2.16)

This solution is in good agreement with the solution obtained using the Lindstedt-poincare’ procedure. [ Nayfeh and Mook, 1979].

Example 6.2.1: Taking α𝟏 = ω𝟎𝟎 = k=100, 2 1α = and 3 1.5α = in eqn 6.2.1find the response of the system. Using Eq. (6.2.16) the variation of frequency with amplitude is shown in figure (6.2.1). Taking two values of a (viz., a=0.009 and a=.029) the time response has been plotted in figure (6.2.2). It may be noted



Figure 6.2.1: Variation of amplitude with frequency for

Figure 6.2.2.(a): Time response (b) Phase portrait corresponding to initial amplitude a=0.009 and a=.029

By changing the quadratic nonlinear terms 2α from .5 to 2.5 and keeping all other parameter same figure (6.2.3) shows the variation of the frequency with amplitude of oscillation.



Figure 6.2.3: Variation of frequency with amplitude for different values of 2α

Figure 6.2.4: Variation of frequency with amplitude for different values of 3α

2 .5α =

2 1.5α = 2 1α =2 2.5α = 2 2α =

3 .5α =3 1α = 3 1.5α = 3 2α = 3 2.5α =



In Figure 6.2.3, by varying the amplitude form -1 to +1 , the frequency of the system changes and it decreases from 1.005 to .98 by increasing 2α from .5 to 2.5. One may observe the reverse phenomenon

by increasing 3α . Also the orientation of the frequency response curve changes from left to right when

3α is greater than 1 or 2 1α ≤ . It may be noted that for a linear system the response frequency of the system does not depend on the amplitude of the response. But in case of the nonlinear system it depends on the response amplitude. A Matlab code is given below which may be used to obtain the frequency response, time response, and phase portrait for different system parameters.

%NPTEL WEB MODULE6 L2--Free Vibration: Duffing Oscillator % Response plot using the method of multiple scales clear all clc omega=1; alpha1=omega^2; alpha2=1; alpha3=1.5; ep=0.1; p1=(9*alpha3*alpha1-10*alpha2^2)/(24*alpha1^2) i=1 for a=0:0.001:1 om=sqrt(alpha1)*(1+p1*ep^2*a.^2) s(i,1)=a; s(i,2)=om; i=i+1; end figure(1) plot(s(:,2),s(:,1),s(:,2),-s(:,1),'linewidth',2) grid on set(gca,'FontSize',15) % For changing fontsize of tick no xlabel('\bf frequency','Fontsize',15) ylabel('\bf amplitude','Fontsize',15) n1=length(s) bt0=0; for ii=10:20:40 om1=s(ii,2) a1=s(ii,1) T=2*pi/om1; jj=1; for t=0:T/1000:10*T p2=ep^2*a1^2*alpha2/(2*alpha1); p3=1-(1/3)*cos(2*om1*t+2*bt0); x(jj)=ep*a1*cos(om1*t+bt0)-p2*p3; tt(jj)=t; jj=jj+1; end n2=jj-1 for k=2:n2 xt(k-1)=(x(k)-x(k-1))/(tt(k)-tt(k-1));



end figure(2) plot(tt,x,'linewidth',2) hold on grid on set(gca,'FontSize',15) % For changing fontsize of tick no xlabel('\bf Time','Fontsize',15) ylabel('\bfx','Fontsize',15) figure(3) plot(x(1:n2-1),xt,'linewidth',2) hold on end grid on set(gca,'FontSize',15) % For changing fontsize of tick no xlabel('\bf displacement','Fontsize',15) ylabel('\bf velocity','Fontsize',15) Exercise problem 6.2.1:

Find the frequency response of a single degree of freedom system with mass=1 kg, stiffness=100 N/m, nonlinear quadratic and cubic stiffness parameter equal to 20 N/m2 and 10 N/m3 respectively. Vary the book-keeping parameter and study the variation of frequency with amplitude.

Module 6 Lecture 3

FREE VIBRATION OF NONLINEAR SINGLE DEGREE OF FREEDOM NONCONSERVATIVE SYSTEMS In this lecture discussion on the vibration of a linear single degree of freedom system with viscous, Coulomb damping, quadratic dumping and will be carried out using method of averaging. System with viscous damping Let us consider a single degree of freedom system with viscous damping. The equation of motion of this system with mass m , stiffness k and damping c can be written as

0mu ku cu+ + = . (6.3.1) The same system can be written using the term natural frequency nω , damping ratioζ as

2 2 0n nu u uω ζω+ + = (6.3.2)

( )2Or, , 2 2n nu u f u u u uω ζω εµ+ = = − = − (6.3.3)



By using Krylov-Bogoliubov method of averaging for an under damped system ( 1ζ < ) the solution can be written as

sin( )nu a tω β= + (6.3.4) where

( )( )2

0

sin cos , sin2 n

n

a f a a dπε φ φ ω φ φ

πω= − −∫ (6.3.5)

( )( )2

0

cos cos , sin2 n

n

f a a da

πεβ φ φ ω φ φπω

= − −∫ (6.3.6)

Substituting expression for ( ),f u u from Eq. (6.3.3) in Eq. (6.3.5) and Eq. (6.3.6), one obtains 2

2

0

sinaa d aπεµ φ φ εµ

π= − = −∫ (6.3.7)

and sin cos 0dεµβ φ φ φπ

= − =∫ . (6.3.8)

Solving Eq. (6.3.7) and Eq. (6.3.8) yields

( ) ( )0 0 0exp exp ,na a t a tεµ ζω β β= − = − = (6.3.9)

Here 0a and 0β are the initial displacement and phase of the response. Substituting Eq. (6.3.9) in Eq.

(6.3.4) one obtains the following equation.

( ) ( ) ( )0 0exp cosn nu a t t Oζω ω β ε= − + + (6.3.10)

This equation is same as the expression one may obtain by finding the complementary function of the differential equation (6.3.2). Using the 0u and 0u as the initial displacement and velocity respectively, one may write Eq. (6.3.10) as

( ) ( )( )0 0 0 sexp cos i/ nn d n d du t u t u u tζω ω ζω ω ω = − + + (6.3.11)

Where the damped natural frequency 21d nω ω ζ= −

For over damped ( 1ζ > ) system one may use the following expressions for the response.

( )( ) ( ) ( )( )( ) ( ) ( )

2 2 2

0 0

2 2 2

0 0

/ 2 exp

/ 2 ex

1 1 1

1 p1 1n n n

n n n

u u u t

u u t

ζ ζ ω ω ζ ζ ζ ω

ζ ζ ω ω ζ ζ ζ ω

= ++ + − − − +

− +

−

− + − − − − −

(6.3.12)

For critically damped ( 1ζ = ) system one may write the response as follows.

( )( ) ( )0 0 0 expn nu u u u t tω ω= + + − (6.3.13)



A Matlab code is given below to plot the under damped, critically damped and over damped response of a system as shown in figure 6.3.1.

Figure 6.3.1(a): Time response of a linear single degree of freedom with viscous damping.

Figure 6.3.1(b): Time response of the system with linear damping. ( )0 03, 1, 0.5, 0.09, 3.15na ω ε µ β= = = = = − Using Eq. (6.3.10) the time response is shown in Fig. 6.3.1(b). It may be noted that the response decreases exponentially. The corresponding Matlab code is given below

Under damped

Critically damped

Over damped



Matlab code 6.3.1: %Free Vibration response of a linear single degree of freedom system x0=0.1; xt0=0.001; wn=2; zeta=1.5; t=0:0.001:20; %overdamped z1=-zeta+sqrt(zeta^2-1); z2=-zeta-sqrt(zeta^2-1); z3=2*wn*sqrt(zeta^2-1); A=(xt0-z2*wn*x0)/z3; B=(-xt0+z1*wn*x0)/z3; x1=A*exp(z1*wn*t)+B*exp(z2*wn*t); %critically damped x2=(x0+(xt0+wn*x0)*t).*exp(-wn*t) %underdamped zt=0.2 %Damping factor wd=wn*sqrt(1-zt^2); x3=exp(-zt*wn*t).*(((xt0+zt*wn*x0)/wd).*sin(wd*t)+x0*cos(wd*t)); plot(t,x1,'r',t,x2,'b',t,x3,'g','linewidth',2) grid on set(gca,'FontSize',15) % For changing fontsize of tick no xlabel('\bf Time','Fontsize',15) ylabel('\bf x','Fontsize',15) Matlab code 6.3.2: % plotting of linear damping (Eq. 6.3.10). clc clear all a0=3; ep=.5; mu=.09; t=0:0.1:100; omega=1; beta=-3.15; a=a0*exp(-ep*mu*t); u=a0*exp(-ep*mu*t).*cos(omega*t+beta); plot(t,u,t,a,'--',t,-a,'--') % title('SYSTEM WITH LINEAR DAMPING') set(findobj(gca,'Type','line'),'Color','b','LineWidth',2); set(gca,'FontSize',14) xlabel('t','fontsize',14,'fontweight','b'); ylabel('u','fontsize',14,'fontweight','b'); grid on



Single degree of freedom system with quadratic damping.

Here, the equation of motion of the system can be written as

( )2 , | |nu u f u u u uω ε+ = = − (6.3.14)

Similar to viscous damping here also using KB method the solution can be written as

sin( )nu a tω β= + (6.3.15)

Here, a and β can be given by Eq. (6.3.5) and (6.3.6). Now using the expression for ( ),f u u in Eq. (6.3.5) and (6.3.6), one may write

( )( )2 22

2

0 0

sin cos , sin sin sin2 2

nn

n

aa f a a d dπ πε ωε φ φ ω φ φ φ φ φ

πω π= − − = −∫ ∫

=2 2

3 3 2

0

4sin sin2 3

nn

a d d aπ π

π

ε ω φ φ φ φ ε ωπ π

− − = − ∫ ∫ (6.3.16)

and ( )( )2 2

0 0

cos cos , sin sin cos sin 02 2

nn

n

af a a d da

π πεωεβ φ φ ω φ φ φ φ φ φπω π

= − − = − =∫ ∫ (6.3.17)

Solving Eq. (6.3.16) and Eq. (6.3.17) one may write Eq. (6.3.15) as ( ) ( )0

00

cos413

nn

au t Oa tω β εεω

π

= + ++

(6.3.18)

Figure 6.3.2: Time response of the system with quadratic damping. ( )0 02, 1, 0.1, 3.15na ω ε β= = = = −



Using Eq. (6.3.18) the time response is shown in Fig. 6.3.2. It may be noted that unlike the linear system the response does not decreases exponentially but decreases algebraically. The corresponding Matlab code is given in Matlab code 6.3.3. Matlab code 6.3.3: % plotting of quadratic damping. (Eq. 6.3.18) clc clear all a0=2; ep=.1; t=0:0.1:80; omega=1; beta=-3.15; a=a0./(1+(4*ep*omega*a0*t)/(3*pi)); u=a.*cos(omega*t+beta); plot(t,u,t,a,'--',t,-a,'--') % title('SYSTEM WITH QUADRATIC DAMPING') set(findobj(gca,'Type','line'),'Color','b','LineWidth',2); set(gca,'FontSize',14) xlabel('t','fontsize',14,'fontweight','b'); ylabel('u','fontsize',14,'fontweight','b'); grid on System with Coulomb damping In this case the equation of motion of the system can be given by

0 for 0

sgn( ) for 0

c

c

mx kx FN x

F N xN x

µµ

µ

+ + =

>= = − <

(6.3.19)

Using method of averaging this equation can be written as

20

for 0/

for 0c

g xx x f F m

g xµ

ωµ− >

+ = = − = <

(6.3.20)

The solution of the above equation can be written as 0cos( )x a tω β= +

where

( )( )2 2

0 0

2sin cos , sin sin sin2 2n

n n n

g ga f a a d d dπ π π

π

ε εµ εµφ φ ω φ φ φ φ φ φπω πω πω

= − − = − − = − ∫ ∫ ∫ (6.3.21)

( )( )2 2

0 0

cos cos , sin cos cos 02 2n

n n

gf a a d d da

π π π

π

ε εµβ φ φ ω φ φ φ φ φ φπω πω

= − − = − − = ∫ ∫ ∫ (6.3.22)

Integrating Eq. (6.3.21) and Eq. (6.3.22) one may obtain



0 0

2 ,n

ga a tπµ β βπω

= − = (6.3.23)

Substituting Eq. (6.3.23) in Eq. (6.3.20) the response of the system can be written as

( ) ( )0 0

2 cos n

n

gx a t t Oπµ ω β επω

= − + +

(6.3.24)

Figure 6.3.3: Time response of the system with Coulomb damping. ( )0 02, 1.3, .1, 0.4, 3.15na gω ε µ β= = = = = −

Here it may be noted that the response of the system decreases linearly. A Matlab code is given below. Matlab code 6.3.4: % plotting of time response for system with Coulomb damping. (Eq.6.3.24) clc clear all a0=2; ep=.1; mug=.4; t=0:0.1:100; omega=1.3; phi=omega; beta=-3.15; a=a0-((2*ep*mug*t)/(pi*omega)); u=a.*cos(omega*t+phi); plot(t,u,t,a,'--',t,-a,'--') % title('SYSTEM WITH COULOMB DAMPING') set(findobj(gca,'Type','line'),'Color','b','LineWidth',2); set(gca,'FontSize',14) xlabel('t','fontsize',14,'fontweight','b'); ylabel('u','fontsize',14,'fontweight','b'); grid on



Exercise problem : 1. Find the response of a single degree of freedom system with mass 1 kg, stiffness 100 N/m and damping factor 10 N.s/m. Plot the time response and phase portrait. Also plot the phase portrait considering coulomb damping and quadratic damping. Develop a Matlab code for finding the time response and phase portrait by using second order governing differential equation of motion (Use Runge-Kutta method). Hints-The Matlab code for the system with viscous damping is given below Matlab code 6.3.5: %Use Runge-Kutta method to obtain the response of a sdof vibrating system m=input( ‘mass of the system in kg = ’ ) k=input( ‘Stiffness of the system in N/m = ’ ) c=input( ‘damping factor of the system in N.S/m= ’ ) u0=input(‘initial Displacement in m= ’ v0=input(‘initial velocity in m= ’ omega_n=sqrt(k/m), zeta=c/(2*m*omega_n); if (zeta>1) display(over damp system) u= end if(zeta==1) display(‘critically damped system’) u= end if(zeta<1) display(‘under damped system’) u=u0*sin(omega_n*t)+(v0/omega_n)*cos(omega_n*t) end plot(t,u) [T,u]=ode45(@ex631f,[0,20],[0.1,0.01] Xlabel Ylabel Title function Ex631f



2. Find the response of a single degree of freedom system with Hysteretic damping. The equation of motion in this case is given by.

2

0x x fω ε+ =

2 and 2

s b b c

s c d

s d a d

s d a

c b s a d s

x x x x x xx x x xf x x x x x x

x x x x

x x x x x x

+ − ≥ ≥ − ≥ ≥− = − − ≥ ≥ ≥ ≥

= − = +

3. Find the response of a single degree of freedom system with material damping by considering (a) Maxwell model (spring and dashpot) in series, (b) Kelvin-Viogot Model (spring and dashpot) in parallel. Consider soft spring with cubic nonlinearity in both the cases.

Module 6 Lecture 4

FREE VIBRATION OF SYSTEMS WITH NEGATIVE DAMPING In this lecture initially the system with negative damping will be discussed. Then the free vibration response of systems similar to van der Pol type of oscillator will be discussed with the help of numerical examples. Finally the nonlinear response of a simple pendulum with viscous damping will be illustrated. There are many systems which can be modeled as a system with negative damping. This type of system particularly occurs in control system where the derivative gains if not properly adjusted will give rise to negative damping. Also this type of damping can be found in the high voltage transmission lines. The equation of motion for this type of system for example that of a Rayleigh oscillator can be given by

( )2 3

0u u f u uω ε ε+ = = − (6.4.1) There are many systems which can be modeled as a system similar Rayleigh oscillator. Using KB method the response amplitude a and phase β of the system can be given by



( )( ) ( )2 2

2 2 2 4

0 0

sin cos , sin sin sin2 2n n

n

aa f a a d a dπ πε εφ φ ω φ φ φω φ φ

πω π= − − = −∫ ∫

2 21 312 4 na aε ω = − −

(6.4.2)

( )( ) ( )2 2

2 2 2

00 0

cos cos , sin 1 sin sin cos 02 2n

n

f a a d a da

π πε εβ φ φ ω φ φ ω φ φ φ φπω π

= − − = − − = ∫ ∫ (6.4.3)

Solving Eq. (6.4.2) one can write

( )

22 0

2 2 2 2

0 0 0 0

3 31 exp4 4

aaa a tω ω ε

= + − −

(6.4.4)

The time responses obtained by using Eq. (6.4.4) are shown in figure 6.4.1 for large and small initial disturbance condition. It is observed that while with large initial disturbance the response decreases to

Figure 6.4.1: Time response of the system with Rayleigh damping. (a) Large initial disturbance ( 0 2a = )

(b) Small initial disturbance ( 0 0.2a = ); ( )01.3, .1, 0.4, 3.15n gω ε µ β= = = = −

attain the steady state periodic response, and in case of small initial disturbance the response grows to attain the steady state periodic response. Matlab code 6.4.1 may be used for plotting the time response of the system with Rayleigh damping. Matlab code 6.4.1: % plotting of time response of the system with Rayleigh damping.Equation no-6.4.4. clc clear all a0=2; %a0=.2; ep=.1; t=0:0.1:80; omega=1;



beta=-3.15; a1=0.75*(omega*a0)^2; a2=a0.^2./(a1+(1-a1)*exp(-ep*t)); a=sqrt(a2); u=a.*cos(omega*t+beta); plot(t,u,t,a,'--',t,-a,'--') % title('SYSTEM WITH Rayleigh DAMPING') set(findobj(gca,'Type','line'),'Color','b','LineWidth',2); set(gca,'FontSize',14) xlabel('t','fontsize',14,'fontweight','b'); ylabel('u','fontsize',14,'fontweight','b'); grid on THE VAN DER POL OSCILLATOR There are many systems which can be modeled as a system similar to van der Pol’s oscillator which is

named after the Dutch physicist Balthasar van der Pol (27 January 1889 – 6 October 1959). Mostly this

equation is used in electrical circuits but it can also be used in some mechanical system where self

oscillation takes place due to negative damping.

The van der Pol’s equation can be written as 2

22 (1 ) d u duu u

dt dtε+ = − (6.4.5)

Using method of multiple scales the solution of this equation can be given by

( ) ( ) ( )0 0 1 1 0 1; , , .......u t u T T u T Tε ε= + + (6.4.6)

Where , 0,1, 2,nnT t nε= = . Using Eq. (6.4.6) in Eq. (6.4.5) and separating the terms with different

order of ε , one obtains the following equations.

20 0 0 0D u u+ = (6.4.7) 2 20 1 1 0 1 0 0 0 02 (1 )D u u D D u u D u+ = − + − (6.4.8)

2 2 2 20 2 2 0 1 1 1 0 0 2 0 0 0 1 0 1 0 0 1 0 02 2 (1 ) (1 ) 2D u u D D u D u D D u u D u u D u u u D u+ = − − − + − + − − (6.4.9)

The solution of Eq.(6.4.7) can be written as 0 0

0 1 2 1 2( , , ) ( , , )iT iTu A T T e A T T e−= + (6.4.10)

Substituting Eq. (6.4.10) in Eq. (6.4.8) 0 032 2 3

0 1 1 1

secular term(2 ] iT iTD u u i D A A A A e iA e cc+ = − − + − +

(6.4.11)

Eliminating the secular term marked in Eq. (6.4.11) one can write



212D A A A A= − (6.4.12)

Now the solution of Eq.(6.4.11) can be written as

0 0331 1 2

1( , )8

iT iTu B T T e iA e cc= + +

(6.4.13)

( )1 2 1 21 ( , ) exp ( , )2

A a T T i T Tφ= (6.4.14)

Substituting Eq. (6.4.14) in Eq. (6.4.12) one can write

( ) ( ) ( ) ( ) ( )2

1 1

1 1 1 1 12 exp exp exp exp exp2 2 2 2 2

a i ia i a i a i a iT T

φφ φ φ φ φ ∂ ∂ + = − − ∂ ∂

(6.4.15)

Separating the real and imaginary terms one can write

2

1 1

1 10, 1 2 4

a a aT Tφ∂ ∂ = = − ∂ ∂

(6.4.16)

Hence, 1

22

2

4( ), and 1 ( ) TT a

c T eφ φ −= =

+ (6.4.17)

So the first order solution of the system can be given by

cos ( )u a t o ε= + (6.4.18) Where

( )2

20

441 1 exp

at

aε

=

+ − −

(6.4.19)

To obtain the second or higher order solution one may use the expression for 0 1 and u u in Eq. (6.4.9) which yields the following equation.

0 020 2 2 1 2 1 2( , , ) ( , , ) NSTiT iTD u u Q T T e Q T T e+ = + + (6.4.20)

Where NST contains non-secular terms.

3 22 2 2

1 2 1 1 12 (1 2 ) 2 (1 2 )8

A AQ iD B i AA B iA B iD A D A AA D A A D A= − + − − − − + − − + (6.4.21)

Now again substituting Eq. (6.4.14) in the secular term of Eq. (6.4.20) and separating the real and

imaginary parts of the resulting equation one obtains the following equation.

32

0, ( )∂= =

∂a a a TT

(6.4.22)



2

1 1 2 1

2 1 7 12 216 16 4

b da d dab a aT a dT dT dT

φ ∂ − = − + + − ∂ (6.4.23)

12

1 7 116 32 8

b dd dT a daa dT a

φ = − + + −

(6.4.24)

Integrating one obtains

31 0 2

2

1 7 1 ln ( )16 64 8

db a T a a a ab TdTφ

= − + + − +

(6.4.25)

In order that the solution to be bounded for all 1T , the coefficient of 1T in the above equation for b must

vanish. Hence one obtains 2

1 016

ddTφ + =

.

So , 2 01

16Tφ φ= − + (6.4.26)

Here 0φ is a constant. Now using the expression for 0 1 and u u the second order solution can be given by

2 20 0

2 20

3 20

7 1 1ln sin 164 8 161cos 1 ( )

16 1 1sin 3 132 16

a a ab tu a t o

a t

ε φε φ ε ε

ε φ

− + − + = − + − + + − +

(6.4.27)

Here a can be given by the Eq. (6.4.19). Equation (6.4.27) can also be written as

3 21cos( ) sin 3( ) ( )32

u a t a t oθ ε θ ε= − − − + (6.4.28)

where 2 20

1 1 7ln16 8 64

t a aθ ε ε ε θ= + − + (6.4.29)



Figure 6.4.2: Time response of the system with van der Pol’s equation ( )01, .1, 3.15nω ε θ= = = −

(a) Small initial disturbance ( 0 0.2a = ) (b) Large initial disturbance ( 0 3.5a = ).

Example: 6.4.2 Simple pendulum with quadratic damping

The equation of motion of a simple pendulum with quadratic damping can be given by

2 2 30 0

12 sin 2 06

θ εµθ θ ω θ θ εµθ θ ω θ θ + + = + + − =

(6.4.30)

Using method of multiple scales with different time scales , 0,1, 2,nnT t nε= = and writing the motion

of the pendulum θ as follows ( ) ( ) ( ) ( ) ...,,,,,,; 2103

32102

22101 +++= TTTTTTTTTt θεθεεθεθ (6.4.31)

and substituting Eq. (6.4.31) in (6.4.29) and separating the terms with different order of ε one obtains the following equations.

01201

20 =+ θωθD (6.4.32)

1102202

20 2 θθωθ DDD −=+ (6.4.33)

3110101

211202103

203

20 6

1222 θθθµθθθθωθ +−−−−=+ DDDDDDDD (6.4.34)

The solution of Eq. (6.4.32) can be written as

( ) ( ) ccTiTTA += 00211 exp, ωθ (6.4.35)

Using Eq. (6.4.35) in (6.4.33) it can be written as



( )( )2 2

0 2 0 2 0 1 1 0 1 0 02 2 expD D D i D A i T ccθ ω θ θ ω ω+ = − = − + (6.4.36)

As the terms in the right hand sides are secular terms, it will be eliminated if 1 0D A = . Hence, A is not a

function of 1T . After eliminating the secular term, the solution of Eq. (6.4.36) will contain only

auxiliary part and hence, 2θ may be dropped. So one can write

( ) ( )1 2 0 0expA T i T ccθ ω= + (6.4.37)

Now substituting (6.4.37) in (6.4.34) one obtains

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( )

' '

0 0 0 0 0 0 0 0 0 0 0

2 2

0 0 0 0 0

2 2

0 3 0 3

3 3

0 0 0 0

0 0 0 0 0

2 exp exp 2 exp exp1 1.exp exp ex

non secular

p exp1 1exp 3 exp 3

2 2

6 6 terms

D

A i T

i A i T A i T i A i T i A i T

i A i T i A i T A A i T AA i T

A i T

ω ω ω µ ω ω ω ω

ω ω ω ω ω

θ ω θ

ω ω

ω

− − − − − + = −

+ +−

+ −

− − +

(6.4.38)

Now substituting ( )βiaA exp21

= in Eq. (6.4.38) one can write

2 2 ' ' 3 3 2 2

0 3 0 3 0 0 0

1 12 sin 2 cos cos cos3 2 sin sin8 24

D a a a a aθ ω θ ω γ ω β γ γ γ µω γ γ+ = + + + + (6.4.39)

where 0tγ ω β= + . It may be noted that the damping term is periodic and can be expanded in Fourier series as

1sin sin sinn

nf nγ γ γ

∞

=

=∑ (6.4.40)

Where 8

3nf π= . The secular terms will be eliminated from () if

016

,03

8

0

2'20' =+=+

ωβ

πµω aaa (6.4.41)

Solving Eq. (6.4.41) one may write

( ) 0200

20

02

200

0

831289

,833

βµωπµω

πβ

µωππ

++

=+

=Ta

aTa

aa (6.4.42)

Hence, the response of the system can be given by



( )( )3

20

0

0020

02

000

0

ˆ1283

ˆ83ˆ1289

cosˆ83

3ε

µωπθ

θωµπµωθπ

ωθωµπ

πθθ O

tt

t+

−

++

+= (6.4.43)

EXERCISE PROBLEM: 1. Obtain the response of a simple pendulum with quadratic damping. The equation of motion for this system can be written as follows. (a) ( ) 2

02 sin 0θ µ θ θ θ ω θ− − + = , (b) ( ) 202 sin 0θ µ θ θ θ ω θ+ − + =

Plot the phase portrait and discuss about the equilibrium solution. 2. Obtain the response of a simple pendulum with viscous damping. The equation of motion for this system can be written as

2 2 30 0

12 sin 2 06

θ µθ ω θ θ µθ ω θ θ + + = + + − =

Plot the phase portrait and discuss about the equilibrium solution. Compare the results with that obtained in problem 6.4.1.

Module 6 Lecture 5

Forced Vibration of single degree of freedom system with cubic nonlinearities



In this lecture, the response of a nonlinear single degree of freedom system with cubic nonlinearies will be discussed considering a weak forcing function. The simplest form of this equation can be given by the forced Duffing equation as follows.

2 30 2 cosu u u u f tω εµ εα ε+ + + = Ω (6.5.1)

As discussed in the previous lectures, in the absence of external force, the free vibration response amplitude of such system is a function of the natural frequency 0ω of the system. Similar to linear vibration of the system here we may consider the behaviour of the system near the resonance condition, i.e., when the external frequency is equal to the natural frequency of the system. This condition is known as the primary resonance condition ( 0ωΩ ≈ ). In case of multi-degree of freedom system one may reduce the system into a number of single degree of freedom system and follow a procedure as outlined here.

To study the behaviour of the system near the primary resonance condition, one may use the detuning parameter which represents the nearness of the external frequency to that of the natural frequency. Hence one may write

0ω εσΩ = + (6.5.2)

Using method of multiple scales the solution of Eq. (6.5.1) can be written as

( ) ( ) ( )0 0 1 1 0 1; , , .......u t u T T u T Tε ε= + + (6.5.3)

Where nnT tε= . Substituting Eq. (6.5.3) in Eq. (6.5.1) and separating terms with different order of ε ,

one obtains the following equations.

2 20 0 0 0D u ω+ = (6.5.4)

2 2 30 1 0 1 0 1 0 0 0 0 0 0 12 2 cos( )D u u D D u D u u f T Tω µ α ω σ+ = − − − + + (6.5.5)


( ) ( ) ( ) ( )0 1 0 0 1 0 0exp expu A T i T A T i Tω ω= + − (6.5.6)

Substituting Eq. (6.5.6) in Eq. (6.5.5) one obtains the following equation.

( ) ( ) ( ) ( )2 2 ' 2 30 1 0 1 0 0 0 0 0 0 0 1

Secular term Mixed Secular term

12 3 exp exp 3 exp2

D u u i A A A A i T A i T f i T T ccω ω µ α ω α ω ω σ + = − + + − + + +

(6.5.7)



To eliminate the secular and near secular terms from Eq. (6.5.7), one can write

( ) ( )20 1

12 ' 3 exp 02

ω µ α σ+ + − =i A A A A f i T .

(6.5.8)

Now substituting ( )1 exp2

A a iβ= in Eq.(6.5.8) and separating the real and imaginary parts following

reduced equations are obtained.

( )10

1' sin 2

µ σ βω

= − + −fa a T

(6.5.9)

( )31

0 0

3 1' cos8 2αβ σ βω ω

= − −fa a T

(6.5.10)

To write these two equations in its autonomous form one may use 1Tγ σ β= − and obtained the following equations.

0

1' sin2

µ γω

= − +fa a

(6.5.11) 3

0 0

3 1' cos8 2αγ σ γω ω

= − +fa a a

(6.5.12)

One should solve these two equations to obtain a and γ and can write the first order solution of the system in the following form

( ) ( )( ) ( )

0 0 1

0

cos ( ) cos ( )

cos ( ) cos ( )

u a t O a t T O

a t t O a t O

ω β ε ω σ γ ε

ω εσ γ ε γ ε

= + + = + − +

= + − + = Ω − + (6.5.13)

Now for steady state as and a γ′ ′equals to 0, one can write Eq. (6.5.11-12) as

0

1 sin2

faµ γω

=

(6.5.14)

3

0 0

3 1 cos8 2

fa aασ γω ω

− = −

(6.5.15)

Now eliminating γ from the above equations, one obtains

2 22 2 2

20 0

38 4

fa aαµ σω ω

+ − =

(6.5.16)



Eq. (6.5.16) is a 6th order Polynomial in a , but quadratic Polynomial in σ . Hence, by solving this quadratic equation, one can write the expression for the frequency response curve as follows.

12 2

2 22 2

0 0

38 4

faa

ασ µω ω

= ± −

(6.5.17)

Hence, for a particular value of detuning parameter one can get different amplitude of the response and

the phase. Now to check the stability of the obtained steady state response, one can perturb the Eq.

(6.5.11) and (6.5.12) to obtain the following Jacobian matrix.

20

00

20

0 0

38

918

aaJ

aa

αµ σω

ασ µω

− − −

=

− −

(6.5.18)

Now to find the stability of the steady state response one can find the eigenvalues by finding the determinant of the J Iλ− matrix. This leads to expression

2 22 2 0 0

0 0

3 92 08 8

a aα αλ µλ µ σ σω ω

+ + + − − =

(6.5.19)

The system will be unstable when the real part of at least one of the eigenvalue becomes positive. This gives rise to the following relation.

2 220 0

0 0

3 9 08 8

a aα ασ σ µω ω

Γ = − − + <



Fig:6.5.1 Frequency response curves for ω=1,f=10, α=10,µ=0.1

Fig:6.5.2 Frequency response curves with different values of α



References for further reading

1. J. A. Goitwald, L. N. Virgin and E. H. Dowell, Experimental mimicry of Duffing's equation,

Journal of Sound and Vibration 158(3), 441461 , (1992)

2. R. E. Mickens, Mathematical and numerical study of the Duffing-harmonic oscillator, Journal

of Sound and Vibration, 244, no.(3), 563–567, 2001

3. S. B. Tiwari, B. N. Rao, N. S. Swamy, K. S. Sai, and H. R. Nataraja, Analytical study on a

Duffing-harmonic oscillator, Journal of Sound and Vibration, 285, (4-5), 1217–1222, 2005.

4. H. Hu, Solution of a quadratic nonlinear oscillator by the method of harmonic balance,

Journal of Sound and Vibration,293, (1-2), 462–468, 2006

5. C. W. Lim, B. S. Wu, and W. P. Sun, Higher accuracy analytical approximations to the

Duffing-harmonic oscillator, Journal of Sound and Vibration, 296, (4-5), 1039–1045, 2006.

6. T. Öziş and A. Yildirim, Determination of the frequency-amplitude relation for a Duffing-

harmonic oscillator by the energy balance method, Computers & Mathematics with

Applications, 54, (7-8), 1184–1187, 2007.

7. A. Beléndez, C. Pascual, E. Fernandez, C. Neipp, and T. Belendez, Higherorder approximate

solutions to the relativistic and Duffing-harmonic oscillators by modified He's homotopy

methods, Physica Scripta,77, Article ID 065004, 14 pages, 2008.

8. J. Cai, X. Wu, and Y. P. Li, An equivalent nonlinearization method for strongly nonlinear

oscillations, Mechanics Research Communications, 32, 553–560, 2005.



Module 6 Lect 6

System with non resonant hard excitations

In the previous lecture a single degree of freedom nonlinear system is considered when the amplit

ude of the external excitation (f) is one order less than the linear term (i.e. 20ω ). In the present lecture

the forcing term is assumed to be of same the order as that of the linear term. So the equation of motion

considered in this case is 2 30 2 cosu u u u f tω εµ εα+ + + = Ω (6.6.1)

Following similar procedure of method of multiple scales, one may write ( ) ( ) ( )0 0 1 1 0 1; , , ...u t u T T u T Tε ε= + + (6.6.2)

Now separating the terms with different order of ε one obtains the following equations. 2 20 0 0 0 0cosD u u f Tω+ = Ω (6.6.3) 2 2 30 1 0 1 0 1 0 0 0 02 2D u u D D u D u uω µ α+ = − − − (6.6.4)


( ) ( ) ( )0 1 0 0 0exp expu A T i T i T ccω= + Λ Ω + (6.6.5)

Where ( )2 202

fω

Λ =−Ω

It may be noted that unlike the previous lecture, where only the complementary part of the solution was

present, in this case both complimentary and particular integral parts are present in the solution of 0u .

Now substituting Eq. (6.6.5) in Eq. (6.6.4) one obtains the following equation.



( ) ( )

( ) ( ) ( )

( )

2 2 ' 2 20 1 0 1 0 0 0

3 3 20 0 0 0 0

20 0

Secular term

Mixed Secular term 1

Mixed

2 6 3 exp

exp 3 ex

Secular ter

p 3 3 exp 2

3 exp

m

2

2

D u u i A A A A A i T

A i T i T A i T

A i T

ω ω µ α α ω

ω ω

αω

+ = − + + Λ +

+ Λ Ω + Λ +Ω

−+ Λ Ω −

( ) ( )

( )

2 20 0 0 0

20

Mixed Secular term 3

Mixed Secular t

3 exp 2 3 exp 2

2 3 6 ex

4

p

erm

A i T A i T

i AA i T cc

ω ω

µ α α

+ Λ + Ω + Λ − Ω −Λ Ω + Λ + Ω +

(6.6.6)

It may be noted from Eq. (6.6.6) that when the exponent terms of the marked mixed secular terms are equal to 0ω a resonance condition will occur. Hence, resonance will be observed in the system when

0

0 0 0

0 0

0 0

(Primary resonance)(Sub harmonic resonan

2 3

13

ce)

(Super harmonic resonance) 3

2 0

ωω ω ω

ω ω

ω ω

Ω =Ω − = ⇒Ω =

Ω = ⇒Ω =

− Ω = ⇒Ω =

For the non resonant case, i.e., when the external frequency is away from 0, 0ω , 013ω or 3 0ω , from Eq.

(6.6.6) eliminating the secular terms yield the following equation.

( )' 2 202 6 3 0i A A A A Aω µ α α+ + Λ + =

(6.6.7)

Now using ( )1 exp2

A a iβ= in Eq.(6.6.7) and separating the real and imaginary parts following reduced

equations are obtained.

1 0 1 0exp( ) exp( )D a a a a T a tµ µ εµ= − ⇒ = − = − (6.6.8)

and 2 2 2 2 2 20 1 1 0 0

1 1 13 3 38 8 8

aD a a a T a tω β α β α β εα β = Λ + ⇒ = Λ + + = Λ + +

(6.6.9)



( ) ( ) ( )

( ) ( )

0 2 20

2 20 0 0 2 2

0

cos cos

1 exp( )cos 3 cos8

fu a t t O

fa t a t t O

ω β εω

εµ ω εα β εω

= + + Ω +−Ω

= − + Λ + + + Ω + −Ω

(6.6.10)

The free oscillation solution decays with time and hence the steady state response consists of forced solution only similar to the linear case.

Superharmonic Resonance ( 013ωΩ ≈ )

To express the nearness of the external excitation frequency to one third of the natural frequency one may use the detuning parameter (σ ) as follows.

03 ω εσΩ = + (6.6.11)

Now to include the mixed secular (or nearly secular or small divisor) term 1in Eq. (6.6.6) in this resonance condition one may write

( )0 0 0 0 0 0 0 0 13 T T T T T Tω εσ ω εσ ω σΩ = + = + = + (6.6.12)

Now to eliminate the secular and near secular terms from Eq. (6.6.6) one can write

( ) ( )' 2 2 30 12 6 3 exp 0i A A A A A i Tω µ α α α σ+ + Λ + + Λ = (6.6.13)

Using ( )1 exp2



( )3

10

sin a a Tαµ σ βωΛ′ = − − − (6.6.14)

( )3

2 21

0 0

3 1 cos8

a a a Tα αβ σ βω ω

Λ ′ = Λ + + −

(6.6.15)

Now to express the above equations in their autonomous form one may use the following transformation.

1Tγ σ β= − (6.6.16)

Hence, Eq. (6.6.14-15) can be written as



3

0

' sina a αµ γωΛ

= − − (6.6.17)

2 3' 3

0 0 0

3 3 cos8

a a aα α αγ σ γω ω ω

Λ Λ= − − − (6.6.18)

By solving the above two equations, one can obtain a and γ , and then can write the solution of the system as

( ) ( ) ( )0 2 20

cos 3 cosfu u a t t Oγ εω

= = Ω − + Ω +−Ω (6.6.19)

For steady state as the time derivative terms vanish, Eq. (6.6.17) and (6.6.18) can be written as

3

0

sina αµ γωΛ

− = (6.6.20)

2 33

0 0 0

33 cos8

a aα α ασ γω ω ω

Λ Λ− − =

(6.6.21)

Now eliminating γ from the above two equations, one can obtain a closed form equation which can be

used for finding the frequency response of the system. 22 2 6

2 2 22

0 0 0

338

a aα α αµ σω ω ω

Λ Λ + − − = (6.6.22)

Solving Eq. (6.6.22) one may write the relation between the detuning parameter and amplitude of the response as follows.

1/22 2 62 2

2 20 0 0

338

aa

α α ασ µω ω ω

Λ Λ= + ± −

(6.6.23)

Hence, in this resonance condition, the free oscillation term does not decay to zero inspite of the presence of damping. Moreover, the nonlinearity adjusts the frequency of the free oscillation term to exactly three times the frequency of the excitations so that the response is periodic. Since the frequency of the free oscillation term is 3 times the frequency of excitation, such resonances are called super harmonic resonances or overtones.

From Eq. (6.6.20) the peak amplitude of the free oscillation term is given by



3

0pa α

µωΛ

= (6.6.24)

2 2 4

2 20 0

3 18p

α ασω µ ω

Λ Λ= +

(6.6.25)

Subharmonic Resonance 03ωΩ =

When the external frequency is nearly 3 times the natural frequency of the system, using detuning

parameter one can write

03 ,ω εσΩ = + (6.6.26)

Or, ( )0 0 0 0 02 T T Tω ω εσΩ − = + (6.6.27)

Using a similar procedure of method of multiple scales, to eliminate the secular terms from Eq. (6.6.6) one can write

( ) ( )' 2 2 20 12 6 3 exp 0i A A A A A A i Tω µ α α α σ+ + Λ + + Λ = .

(6.6.28)

Using ( )1 exp2



( )21

0

3' sin 34

a a a Tαµ σ βωΛ

= − − − (6.6.29)

( )' 2 2 21

0 0

3 1 3 cos 38 4

a a a a Tα αβ σ βω ω

Λ = Λ + + − (6.6.30)

Now to express the above equations in their autonomous form one may use the following transformation.

1 3Tγ σ β= − (6.6.31)

2

0

3' sin4

a a aαµ γωΛ

= − − (6.6.32)



2' 3 2

0 0 0

9 9 9 cos8 8

a a a aα α αγ σ γω ω ω

Λ Λ= − − − (6.6.33)

( ) ( ) ( )12 20

1cos cos3

γ ω− = Ω − + −Ω Ω + ∈

u a t f t O (6.6.34)

For steady state one can write

2

0

3 sin4

a aαµ γωΛ

− = (6.6.35)

23 2

0 0 0

9 9 9 cos8 4

a a aα α ασ γω ω ω

Λ Λ− − =

(6.6.36)

Now eliminating γ from the above two equations, one can obtain a closed form equation which can be

used for finding the frequency response of the system.

22 2 22 2 2 4

20 0 0

9 9 8198 16

a a aα α αµ σω ω ω

Λ Λ + − − = (6.6.37)

This shows either the system will have a trivial state response (i.e., a =0) and a non trivial response which can be obtained by solving the following equation.

22 2 22 2 2

20 0 0

9 9 8198 16

a aα α αµ σω ω ω

Λ Λ + − − = (6.6.38) This equation is quadratic in 2a and hence the solution can be written as

( )1/22 2a p p q= ± − (6.6.39)

Where 208 69

p ω σα

= − Λ (6.6.40)

and 22 2

202

0

64 99 81

q ω αµ σα ω

Λ = + − (6.6.41)



As q is always positive, so the nontrivial free oscillation amplitudes occur when 20 and p p q> ≥ . For these conditions one should have

2 22 20

0 0

4 63 and 2 027 8ω σ α ασ µα ω ω

Λ ΛΛ < − − ≥

(6.6.42)

So σ and α should have same sign. From Eq. (6.6.42), for a given Λ nontrivial solutions can exist only if

2 2 20

20

2 638

µ ω αασωΛ

≥ +Λ (6.6.43)

Similarly for a given σ , nontrivial solution can exist only if

1/2 1/22 2 2

2 20

6363 634

σ σ α σ σµ µ ω µ µ µ

Λ− − ≤ ≤ + − (6.6.44)

In the σΛ plane the boundary of the region where nontrivial solutions can exist can be given by

1/ 22 2

20

63 634α σ σω µ µ µ

Λ= ± −

(6.6.45)

Exercise Problems

1. Study the response of the single degree of freedom system with both quadratic and cubic nonlinearities. Consider primary, subharmonic and superharmonic resonance conditions. Use either method of multiple scales or the method of normal forms.

References:

1. A. H. Nayfeh and D. T. Mook: Nonlinear Oscillations, Wiley, 1979

2. A. H. Nayfeh, Method of Normal Forms, Wiley,1993

Exercise problems

1. 1. Using numerical techniques plot the time response of the system given by Eq. (6.6.1) with primary, subharmonic and superharmonic resonance. Compare the results with those obtained from the perturbation method.

2. Study different resonance conditions for the Duffing equation with two frequency excitation terms as given by the following equation.



( ) ( )2 30 1 1 1 2 2 22 cos cos u u u u f t f tω εµ εα θ θ+ + + = Ω + + Ω +

Ans: One may observe following resonance conditions

Q No. 3 Plot the frequency response curves for the system with sub harmonic and super harmonic resonance condition using equation 6.6.25 and 6.6.45.

Module 6 Lecture 7

Forced vibration Single-Degree of freedom system

In this lecture briefly following analysis will be carried out

• Forced vibration of Single-Degree of freedom system with cubic and quadratic nonlinearies • System with self sustained oscillation

System with cubic and quadratic nonlinearies

In this case considering damping, cubic nonlinearity and the forcing parameters to be one order less than quadratic nonlinearity which is one order less than the linear term, the equation of motion can be written as

2 2 2 2 3 20 2 32 cosu u u u u f t+ + + + = Ω ω ε µ εα ε α ε (6.7.1)

Following similar procedure as in the previous two lectures one may study the primary, super harmonic and sub harmonic resonance conditions.

( )

( )

0 0 0 0

0 1 2 0 1 0 2

0 2 1 1 2 0 2 1 0 2 1

2 1 0 2 1 0

2 1 0 2 1

2 1 0 2 1 0

2 1 0 2 1 0

1 13 , , , 1 23 2

1 13 3 , 3 3

12 2 , 2 ,2

9 3 , 31 5, 53 3

7 27 , 23 3

7 57 , 53 3

n n m n m n

or or

or

ω ω ω ω

ω ω ω

ω ω ωω ω

ω ω

ω ω

ω ω

≈ Ω ≈ Ω ≈ + ± Ω ±Ω ≈ Ω ±Ω

≈ Ω Ω ≈ Ω ≈ Ω

≈ Ω ± Ω Ω −Ω ≈ Ω ±Ω ≈ Ω ±Ω

Ω ≈ Ω ≈ Ω ≈ Ω ≈

Ω ≈ Ω ≈ Ω ≈ Ω ≈

Ω ≈ Ω ≈ Ω ≈ Ω ≈

Ω ≈ Ω ≈ Ω ≈ Ω ≈



In case of primary resonance, taking 20Ω = +ω ε σ and

( ) ( ) ( ) ( ), , , , , , ,20 0 1 2 1 0 1 2 2 0 1 2u t u T T T u T T T u T T T= + +ε ε ε (6.7.2)

and applying the usual procedure of method of multiple scales, one can obtain the following reduced equations (Nayfeh and Mook, 1979).

0

' sin2

fa aµ γω

= − − (6.7.3)

2 2' 33 0 2

30 0

9 10 cos24 2

fa a aα ω αγ σ γω ω−

= − + (6.7.4)

Comparing these two equations with those obtained from Duffing equation with only cubic

nonlinearities for primary resonance (Eq. 6.5.11 and 12), one may observe that both sets of equations

are identical if 22

3 20

109αα αω

= − (6.7.5)

When 22

3 20

1009αα αω

= ⇒ = − which is negative, the system will show softening effect. Also if

22

3 20

10 , 09αα αω

< < one obtains softening effect in which the frequency response curves bends towards the

lower frequencies irrespective of the sign of 2α .

When 22

3 20

10 ,9ααω

= 0α = , the system behaviour will be similar to that of a linear system. Here the effect of

cubic nonlinearities will be cancelled by that of the quadratic nonlinearity. Similar to the expression in the cubic nonlinear system here the following equation can be used to find the frequency response.

2 22 2 2

20 0

38 4

fa aαµ σω ω

+ − =

(6.7.8)

The expression for the response can be given by

( ) ( ) ( )2 2 22 22 20 0

1 1cos cos 2 22 6

u a t a a t Oα αγ ε ε γ εω ω

= Ω − − + Ω + + (6.7.9)



Comparing this expression with that of the system with only cubic nonlinearity, it can be observed that the oscillating motion is not centered at 0u = and there is a drift or shift of the steady state part by an amount of

2220

12

aαεω

− .

Superharmonic resonance

This resonance condition may occur when one consider, the forcing term is same order that of the linear part. Also considering the damping and quadratic nonlinearity of the order of ε and cubic nonlinearity of the order

of 2ε one may write the equation of motion as (Nayfeh and Mook, 1979)

2 2 2 30 2 32 cosu u u u u f tω εµ εα ε α+ + + + = Ω (6.7.10)

Applying method of multiple scales while eliminating the secular term, one may observe that a resonance condition will occur when 02 ω εσΩ = + . In this case the steady state solution can be written as

( )( ) ( )

( ) ( )2

2222 2 1/22 2 2 2

0 0 0

cos sin 24

ffu t t Oα γ εω ω ω µ σ

= Ω − Ω − +−Ω −Ω −

(6.7.11)

1tanσ

γµ

− =

Subharmonic resonance

For the system given by equation (6.7.10), subharmonic resonance will occur when external frequency is nearly

equal to 02ω εσΩ = + . In this case one may obtain the expression for the frequency of resulting oscillation as 1/22 2 2

220 4

α σλ µω

Λ−= − ±

(6.7.12)

Hence if 2 2

2 220

4ασωΛ

> , the motion is oscillatory and decays with time

If 2 22 2

22 222200

44 4 αα µσωω

ΛΛ−> >

, the response decays without oscillating

If 2 2

2 2220

44 α µ σω

Λ− >

, the system becomes unstable as the response grows

Systems with self-sustained oscillations

Let us consider the van der Pol’s oscillator with soft harmonic excitation which is given by the following equation.



320

1 cos3

f tu uu u = Ω−

+ − εω ε (6.7.13)

Let us consider the primary resonance case in which the external frequency is assumed to be near the linear system frequency 0ω . So using detuning parameter one may write

0tΩ = +εσω (6.7.14)

Now to solve Eq. (6.7.14), taking

( ) ( ) ( ) ( ), , ,u t u T T u T T oε ε ε= + + 20 0 1 1 0 1 (6.7.15)

one will obtain the following reduced equations

2 20

0

1 1' sin12 24

fa aa γωω

= +− (6.7.16)

'

0

cos2

fa aγ σ γω

= + (6.7.17)

The steady state solution can be given by ( ) ( )cosu a t Oγ ε= Ω − + (6.7.18)

For steady state taking 'a = 'γ =0 and eliminating γ form the equations (6.7.16-17) one can obtain

( )2

22 414fρ σ ρρ + =− (6.7.19)

Where 2 20

14

aρ ω= (6.7.20)

Figure 6.7.1 shows the frequency response obtained using equation 6.7.20 for different values of forcing parameter.



Fig. 6.7.1: Frequency response curves for primary resonance for van der Pol’s oscillator

Exercise Problems:

1. Using numerical techniques plot the time response of the system given by Eq. (6.7.1) with primary, subharmonic and superharmonic resonance. Compare the results with those obtained from the perturbation method.

2. Find the frequency response curves for van der Pol’s oscillator considering strong forcing term. Study the subharmonic and superharmonic resonance conditions.

3. Find the expressions for frequency response curves for a single degree of freedom system for different resonance conditions when subjected to 2 and 3 frequency excitations.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

σ

ρ

Response of the vander pol oscillator for primary resonance



Module 6 Lecture 8

Free and Forced vibration of Multi-Degree of freedom system

In this lecture briefly we will discuss about the free vibration of multi-degree of freedom nonlinear systems. Initially the system with quadratic nonlinearity and then the system with cubic nonlinearity will be considered.

Free vibration of the system with quadratic nonlinearities

Let us consider a two degree of freedom system where the equation of motion can be given by

21 1 1 1 1 1 1 2 0ˆ2u u u u uω µ α =+ + − (6.8.1)

2 22 2 2 2 2 2 1ˆ2 0u u u uω µ α+ + − = (6.8.2)

To use method of multiple scales one may assume the solution of these equations using different time scales ( )0 1, n

nT T T tε= as follows.

21 11 0 1 12 0, 1( , ) ( ) ...u u T T u T Tε ε= + + (6.8.3)

22 21 0 1 22 0, 1( , ) ( ) ...u u T T u T Tε ε= + + (6.8.4)

Substituting Eq. (6.8.3) and (6.8.4) in Eq. (6.8.1) and (6.8.2) and separating the terms with different order of ε one obtains

Order of ε

2 20 11 1 11 0D u uω+ =

(6.8.5)

2 20 21 2 21 0D u uω+ =

(6.8.6)

Order of 2ε

2 20 12 1 12 0 1 11 1 11 1 11 212 ( )D u u D D u u u uω µ α+ + − + + (6.8.7)

2 2 20 22 2 22 0 1 21 2 21 1 112 ( )D u u D D u u uω µ α+ + − + + (6.8.8)



The solution of Eq. (6.8.5) and (6.8.6) can be given by

11 1 1 1 0( ) exp ( )u A T i T ccω= + (6.8.9)

21 2 1 2 0( ) exp ( )u A T i T ccω= + (6.8.10)

Substituting Eq. (6.8.9) and (6.8.10) in Eq. (6.8.7) and (6.8.8) yields

[ ] 2 20 12 1 12 1 1 1 1 1 1 0 1 1 2 1 2 0

Secul2

ar term( )exp ( ) exp ( )D u u D A A i T A A i T ccω ω µ ω α ω ω+ = − + + + +

(6.8.11)

2 2 20 22 2 22 2 1 2 2 2 2 0 2 1 1 0 1 1

Secular 2 ( ) exp ( ) exp(2

e m)

T rD u u D A A i T A i T A A ccω ω µ ω α ω + = − + + + +

(6.8.12)

To eliminate the secular term from Eq. (6.8.11) and Eq. (6.8.12) one can write

1 1 1 1 0D A Aµ+ = and 1 2 2 2 0µ+ =D A A (6.8.13)

Solving Eq. (6.8.13) one can write

1 1 1 1exp( )A a Tµ= − and 2 2 2 1exp( )A a Tµ= − (6.8.14)

Substituting Eq. (6.8.14) in (6.8.3) and (6.8.4), the first order solution of the system can be given by

( ) ( )21 1 1 1exp( ) expu t a i t cc Oε εµ ω ε= − + + (6.8.15)

( ) ( )22 2 2 2exp( ) expu t a i t cc Oε εµ ω ε= − + + (6.8.16)

Hence, for steady state as time tends to infinity, both the response 1 2 0u u= = .

Resonant case (System with internal resonance)

Considering internal resonance of 1:2, i.e., when the second frequency is nearly equal to twice the first frequency one can write

2 12ω ω εσ= + (6.8.17)

So, 1 0 2 0 0 2 0 12 T T T T Tω ω εσ ω σ= − = − and ( )2 1 0 1 0 0 1 0 1ω ω ω εσ ω σ− = + = +T T T T T (6.8.18)

Following similar procedure of method of multiple scales to eliminate the secular terms one obtains the following equations.



( ) ( )'1 1 1 1 1 2 1 12 exp 0i A A A A i Tω µ α σ− + + = and ( ) ( )' 2

2 2 2 2 2 1 12 exp 0i A A A i Tω µ α σ− + + − = (6.8.19)

Now using ( ) ( )1 1 1 2 2 21 1exp and exp2 2

A a i A a iθ θ= = and introducing 2 1 12 Tγ θ θ σ= − + and separating

the real and imaginary parts the following reduced equations are obtained.

' 11 1 1 1 2

1

sin4

a a a aαµ γω

= − + and ' 222 2 2 1

2

sin4αµ γω

= − −a a a (6.8.20)

' 11 1 1 2

1

cos4

a a aαθ γω

= − and ' 222 2 1

2

cos4

a aαθ γω

= − (6.8.21)

Or, 2 2

1 2 2 12 2

1 2

cos2 4

a aa a α αγ σ γω ω

′ = + −/

(6.8.22)

For steady state using ' ' '1 2 0a a γ= = = , one may write

11 1 1 2

1

sin 04

a a aαµ γω

− + = (6.8.23)

222 2 1

2

sin 04

a aαµ γω

− − = (6.8.24)

2 21 22 1 2

1 2

cos 02 4

a a aα α γ σω ω

− + =

(6.8.25)

Eliminating γ from Eq. (6.8.23) and (6.8.24) one obtains the following equations.

2 22 2 11 2

1 1 2

0a aµ ω αµ ωα

+ = (6.8.26)

If 1 2 and α α are of different sign, 1 2 and a a can differ from zero, that is the system may have nontrivial response. Hence, in the presence of internal resonance, even though there is no external forcing energy will transfer from 1st mode to second mode and self sustained oscillation will continue.

Forced vibration Multi-Degree of freedom system with quadratic nonlinearities

Let us consider a two degree of freedom system with quadratic nonlinearies given by the following equations.



( )1 12

1 1 1 1 1 1 1 2 cos2 f tu u u u u ε τω εµ α = Ω ++ + + (6.8.27)

( )2 2 22 2 2 2 2 2 1 2 22 cosu u u u f tω εµ α ε τ+ + + = Ω + (6.8.28)

Now to solve these equations let us use method of multiple scales by considering

21 11 0 1 12 0, 1( , ) ( ) ...u u T T u T Tε ε= + + and

22 21 0 1 22 0, 1( , ) ( ) ...u u T T u T Tε ε= + + (6.8.29)

Substituting Eq. (6.8.29) in Eqs. (6.8.27-6.8.28) and equating coefficients of like power of ε one obtains

( )2 20 11 1 11 1 0 1cosD u u f Tω τ+ = Ω + (6.8.30)

2 20 21 2 21 0D u uω+ = (6.8.31)2 20 12 1 12 0 1 11 1 11 1 11 212 ( )D u u D D u u u uω µ α+ = − + − (6.8.32)

2 2 20 22 2 22 0 1 21 2 21 2 11 2 0 22 ( ) cos( )D u u D D u u u f Tω µ α τ+ = − + − + Ω + (6.8.34)

Now, solution of Eq. (6.8.30) and Eq. (6.8.31) can be given by

( )11 1 1 1 0 0 1( ) exp ( ) expu A T i T i T ccω τ= +Λ Ω + + (6.8.35)

21 2 1 2 0( ) exp ( )u A T i T ccω= +

(6.8.36) Here, ( )2 21 1/ 2f ωΛ = −Ω Now

substituting these two equations in Eq. (6.8.33) and (6.8.34) one can write( )2 2 '

0 12 1 12 1 1 1 1 1 0 1 2 1 2 1 0

secular term

2 ( )exp ( ) expD u u A A i T A A i Tω ω µ ω α ω ω+ = − + − +

( ) ( )2 1 2 1 0 2 2 0 1

122 1near secular in case of internal resonance

exp expA A i T A i T iω ω

ω ω ω τα

≈

− + Λ Ω+ + −

( ) ( )1

1 2 2 0 1 1 0 1

1 2near secular term if near secular term if

exp 2 expA i T i i i T ccω ω ω

α ω τ µ τ+Ω ≈ Ω ≈

− Λ Ω− + − ΩΛ Ω + +

(6.8.37)

And ( )2 2 ' 2

0 22 2 22 2 2 2 2 2 0 2 1 1 0

Secular term

2 ( )exp ( ) exp 2D u u A A i T A i Tω ω µ ω α ω+ = − + −

( )( )22 1 1 1 1 0 12 expA A A i T iα ω τ− +Λ + Λ +Ω +



( ) ( )2

1 1 0 1 0 12

may be near secular term considering internal resonance

2 exp exp 2A i T i i Tω τ τα

Λ Ω− + +Λ Ω + −

( )

2

2 0 2

near secular term if

1 exp2

f i T cc

ω

τ

Ω≈

+ Ω + +

(6.8.38)

From Eq. (6.8.37) and (6.8.38) one may observe that one may get many resonance conditions such as (a) 1ωΩ ≈ , (b) 2ωΩ ≈ (c) 1 2ω ωΩ ≈ + . Also, some resonance condition occurs when one consider

internal resonance i.e. 2 12ω ω≈ .

Let us consider when the external frequency is nearly equal to the second mode frequency i.e.,

2 1ω εσΩ = + (6.8.39).

Hence, without considering internal resonance, to eliminate secular terms one can write

'1 1 1 0A Aµ+ = (6.8.40)

and ( ) ( )'2 2 2 2 2 1 1 2

12 exp2

i A A f i Tω µ σ τ+ = + (6.8.41)

Now solution of Eq. (6.8.40) can be written as

1 1 1 1 11 exp( )2

A a T iµ θ= − + (6.8.42)

( ) ( )22 2 2 1 2 1 1 2

2 2 1

1 exp( ) exp2 4

ifA a T i i Ti

µ θ σ τω µ σ

= − + − + + (6.8.43)

For steady state as t tends to 0, one obtains

( ) ( )21 2 1 1 2

2 2 1

0, and exp4

ifA A i Ti

σ τω µ σ

= = − + + (6.8.44)

Hence one obtains,

( ) ( ) ( )211 12 2

1

cosFu t Oτ εω

= Ω + +−Ω

(6.8.45)



( )( ) ( )22

2 2 02 22 2 1

sin2

Fu t Oτ γ εεω µ σ

= Ω + − ++

(6.8.46)

where

1 10

2

tanσ

γµ

− =

(6.8.47)

Now if one considers the system with internal resonance i.e., 2 12ω ω≈ or 2 1 22ω ω εσ= + , then to eliminate secular term one can write

( ) ( )' '1 1 1 1 1 2 1' 2 12 exp 0i A A A A i Tω µ α σ− + − = (6.8.48)

( ) ( ) ( )' 22 2 2 2 2 1 2 1 2 1 1 2

12 exp exp 02

i A A A i T f i Tω µ α σ σ τ − + − + − + = (6.8.49)

Substituting ( )1 exp2n n nA a iβ= where n =1, 2, in Eqs. (6.8.48) and (6.8.49) one obtains

' 11 1 1 1 2 2

1

sin4

a a a aαµ γω

= − − (6.8.50)

' 2 122 2 2 1 2 2 2 1

2

1sin sin4 2

a a a fαµ γ ω γω

−= − + + (6.8.51)

' 11 1 2 1 2

1

cos4

a a aαβ γω

= (6.8.52)

' 222 2 1 2 2 1

2 2

1cos cos4 2

a a fαβ γ γω ω

= − (6.8.53)

Where, 1 1 1 2 2Tγ σ β τ= − + and 2 2 1 2 12 Tγ β β σ= − − (6.8.54)

For steady state response, '1a , '

2a , '1β '

2β are zero and there are two possibilities. In the first case one will obtain the linear solution similar to Eq. 6.8.44. In the second case

1/21/22 2

1 1 2 2124

a P f P = − ± −

(6.8.55)



( )1/222

2 1 1 1 22 4a ω µ σ σ = + − (6.8.56)

Here, ( )1 1 2 1 2 1 1 22 2P ωω σ σ σ µ µ = − +

( )2 1 2 1 1 2 2 12 2P ωω σ µ µ σ σ = − − (6.8.57)

Substituting these values in 1u and 2u one can obtain the response of the system. It may be noted that the

second mode amplitude 2a is not a direct function of the external excitation. Using these equations one may plot the frequency response curves and observe many resonance phenomena similar to those observed in case of single degree of freedom system.

Exercise Problems:

1. Plot the frequency response curves using Eq. (6.8.55) and (6.8.56) taking 1α = 2α =-1. Observe different nonlinear phenomena. Study the stability of the system.

2. Study the resonant and non resonant free vibration of two degree of freedom system with cubic nonlinearities. The equation of motion of this system can be given by

2 3 2 2 31 1 1 1 1 1 1 2 1 2 3 1 2 4 2ˆ2 0u u u u u u u u uω µ α α α α+ + + + + + =

2 3 2 2 31 1 1 1 1 1 2 1 2 3 1 2 4 2ˆ2 0u u u u u u u u uω µ α α α α+ + + + + + =



Module 6 Lect 9

Parametrically excited system

In this lecture, the parametrically excited system will be considered and using Floquet theory the conditions for instability regions will be determined. It may be noted that a simple parametrically excited system can be represented by the following equation.

1 2( ) ( ) 0u p t u p t u+ + = (6.9.1)

Where, the term 1( )p t and 2 ( )p t are periodic function of time. It may be noted that as these time varying terms are coefficients of the response and its derivative, this equation is called the equation of a parametrically excited system. One can have many variation of this equation by including different nonlinear terms and forcing terms. One can consider a single or multi degree of freedom system also. Eq. (6.9.1) can also be written in the following form by substituting

11exp ( )2

u x p t dt = − ∫

in Eq. (6.9.1). The resulting equation can be written as

( ) 0x p t x+ = (6.9.2)

where 22 1 1

1 1( )4 2

p t p p p= − − (6.9.3)

Equation (6.9.2) is called the Hill’s equation who studied this system in 1886 (Nayfeh and Mook 1979).

Now by substituting ( ) 2 cos 2p t t= +δ ε in Eq. (6.9.2) one can write

( 2 cos 2 ) 0x t x+ + = δ ε (6.9.4)

This equation is known as Mathieu’s equation. It may be noted that the basics of parametrically excited systems are based on Mathieu-Hill types of equation. As pointed out in module 4, one may use Floquet theory to study the stability of these systems.



Example 6.9.1: Study the stability of the Hill’s equation given in Eq. (6.9.2) by taking the following initial condition.

1 1 2 2(0) 1, (0) 0, (0) 0, (0) 1u u u u= = = = (6.9.5)

Solution

Writing the fundamental sets of solution as

1 11 1 12 2

2 21 1 22 2

( ) ( ) ( )( ) ( ) ( )

u t T a u t a u tu t T a u t a u t

+ = ++ = +

(6.9.6)

1 11 1 12 2

2 21 1 22 2

( ) ( ) ( )( ) ( ) ( )

u t T a u t a u tu t T a u t a u t

+ = ++ = +

(6.9.7)

one can obtain

11 1 21 2 12 1 22 2( ), ( ), ( ), ( )a u T a u T a u T a u T= = = = (6.9.8)

Or 1 1

2 2

( ) ( )( ) ( )

u T u TA

u T u T

=

(6.9.9)

Finding the determinant of A I−λ matrix one may write

2 2 0λ αλ− + ∆ = (6.9.10)

where

11 2 1 2 1 22 [ ( ) ( )], ( ) ( ) ( ) ( )u T u T u T u T u T u T= + ∆ = − α (6.9.11)

The parameter ∆ is known as the Wronskian determinant of 1( )u T and 2 ( )u T .

In case of Hill’s equation the Wronskian determinant can also be obtained as follows.

As, 1( )u t and 2 ( )u t are the fundamental set of solution, hence they must satisfy Eq. (6.9.2). Hence, one can

write

1 1

2 2

( ) 0( ) 0

u p t uu p t u+ =+ =

(6.9.12)

Or, 1 2 1 2 0u u u u− = (6.9.13)

Which can be integrated to obtain, 1 2 1 2( ) ( ) ( ) ( ) ( )t u t u t u t u t∆ = − =constant (6.9.14)



At t =0, ( )t∆ = 1. So the roots of the Eq. (6.9.10) can be given by

21,2 1λ α α= ± − (6.9.15)

Or, 1 2 1λ λ = (6.9.16)

From Eq.(6.9.15) it may be noted that α >1 one of the root will be greater than unity while the other root is

less than one. Hence one of the normal solutions is unbounded and the other is bounded. When α <1 both the roots will be complex conjugate and their absolute value will be less than one. Hence they will be in the unit circle. So the solutions will be bounded. It may be observed that the transistion from stable to unstable will

takes place when α =1. This corresponds to a periodic solution of period T when 1 2 1= =λ λ and a periodic

solution of period 2T when 1 2 1= = −λ λ .

Example 6.9.2: Study the stability of the Mathieu’s equation using same initial condition given in example 6.9.1.

Solution: In case of Mathieu equation

( 2 cos 2 ) 0x t x+ + = δ ε (6.9.17)

α is a function of δ and ε . The values of δ and ε for which α >1 are called unstable values while those for

which α =1 are called transition values. The locus of transition values separates the ε δ plane into regions of

stability and instability as shown in the fallowing figure.

Fig 6.9.1 Transition curve for Matheiu equation

Hill’s Infinite Determinat

1.9 2 2.1 2.2 2.3 2.4 2.50

0.5

1

1.5

2

ε

ω

Stable Stable

Unstable



One may also use hill’s infinite determinant method to find the transition curve which is explained below.

( )2 cos 2 0u t u+ δ + ε = (6.9.18)

Using Floquet theory one may assume the solution of the equation (6.9.18) as

( ) ( )expu t t= γ φ (6.9.19)

Where ( ) ( )t t Tφ = φ + . One may expand ( )tφ in a Fourier series to obtain the following equation.

( )exp 2nn

u in t∞

=−∞

= φ γ + ∑ (6.9.20)

Where nφ is constant. Substituting Eq. (6.9.20) in (6.9.18) one obtains

( ) ( ) ( ) ( ) 22 exp 2 exp 2 1 exp 2 1 0n nn n

in in t t i n t t i n t∞ ∞

=−∞ =−∞

γ + + δ φ γ + +∈ φ γ + + + γ + − = ∑ ∑ (6.9.21)

Equating each of the coefficients of the exponential functions to the zero one can obtain the following infinite

set of linear, algebraic, homogenous equations for nφ

( ) ( )2

1 12 0n n nin − + γ + + δ φ + ε φ + φ = (6.9.22)

For a non trivial solution the determinant of the coefficient matrix must vanish. Since the determinant is infinite one may divide the mth row by 24mδ − for convergence considerations to obtain the following Hill’s determinant.



( )

( )

( )

( )

( )

2

2 2

2

2 2 2

2

2

2 2 2

2

2 2 2

... . . . . . . . . . ...4

... 0 0 0 0 0 0 0 ...4 4

2... 0 0 0 0 0 0 ...

2 2 2... 0 0 0 0 0 0 ...

2... 0 0 0 0 0 0 ...

2 2 24

... 0 0 0 0 0 0 ...4 4 4

i

i

i

i

δ + γ −εδ − δ −

δ + γ −ε εδ − δ − δ −

ε δ + γ ε∆ γ =δ δ δ

δ + γ −ε εδ − δ − δ −

δ + γ −ε εδ − δ − δ −

(6.9.23)

The determinant can be rewritten as (Whittaker and Watson 1962, Nayfeh and Mook 1979)

( ) ( )

−∆=∆δπ

πγγ

21sin

21sin

02

2 i (6.9.24)

Since the characteristic exponents are solution of ( )∆ γ =0, they are given by

( )21

21 621sin0sin2

∆±= − π

πγ i

(6.9.25)

Once γ is known nφ can be related to 0φ using equation (6.9.22).

One may also consider the central three rows and columns to approximate the characteristic equation as follows

( )( )

( )

2

2

2

2 00

0 2

i

i

δ + γ − ε∆ γ = =ε δ + γ ε

ε δ + γ +

Or ( ) ( ) ( ) ( ) ( )2 2 2 22 2 22 2 2 2 0i i i i δ + γ + δ + γ δ + γ − − ε δ + γ + − ε δ + γ − = (6.9.26)

The transition curve separating stability from instability correspond to γ =0 (i.e., periodic solution with periodπ) or iγ = ± (i.e., periodic motion with period 2π ).



When γ =0 Eq. (6.9.26) gives the transition curves

212

δ = − ε and 2142

δ = + ε (6.9.27)

When iγ = ± Eq. (6.9.26) gives the transition curves

1δ = ± ε and 2198

δ = + ε (6.9.28)

Exercise problem

Problem 6.9.1: Use Floquet theory to study the stability of the periodic motion corresponding to primary resonance of Duffing equation.

Problem 6.9.2: Use Method of Multiple Scales to determine the equations for transition curves for Matheiu equation. Plot these transition curves near 2δ = ω =1, 4. Taking few points in the stable and unstable regions and by using numerical method to solve the Mathieu’s equation, plot the time responses to check whether the marked instability regions are correct.

Module 6 Lect 10

Multi-degree-of freedom parametrically excited system

Case study: Instability Region of a Sandwich beam

In this lecture a case study has been taken by considering a three-layered, soft-cored, symmetric

sandwich beam subjected to a periodic axial load. For completeness purpose the derivation of the

governing equation of motion is given here and then the parametric instability regions for simple and

combination resonances are investigated for simply supported and clamped-free end conditions by

modified Hsu’s method.



Figure 6.10.1 Symmetric three-layered soft cored sandwich beam with periodic axial load

Figure 6.10.1 shows a simply supported, symmetric, three-layered sandwich beam of length L and

width b with a flexible soft core. The top, core and bottom layer thickness are d t , c and d b ,

respectively. The upper and lower layers (face layer) of the beam are of the same elastic material and

the core is of soft viscoelastic material. The sandwich beam is subjected to an axial periodic load

( ) ,cos10 tPPtP ω+= ω being the frequency of the applied load, t being the time and 0P and 1P are

the amplitudes of static and dynamic load, respectively.

Figure 6.10.2 shows the geometry of the sandwich beam, the load and internal forces and moments in

different layers and the deflection in x and z directions before and after deformations. Here, xxQ is the

shear force, xxN is the axial force and xxM is the bending moment. Superscripts and t b represent the

top and bottom layer, respectively. The assumptions made for deriving the governing equations are

similar to that by Frostig [1-4] and are (i) the face sheets of the sandwich beam are modeled as Euler-

Bernoulli beams (ii) the transversely flexible core layer is considered as a two dimensional elastic

medium with small deformations where its height may change under loading, and its cross section does

not remain planar. The longitudinal (in-plane) stresses in the core are neglected and (iii) the interface

layers between the face sheets and the core are assumed to be infinitely rigid and provide perfect

continuity of the deformations at the interfaces.

NA

NA



Figure 6.10.2 (a) Geometry, (b) load, internal forces and moments (c) displacement pattern through depth of

section. N.A is the neutral axis.

The internal potential energy (U ) in terms of direct stresses (σ) and shear stress τ and strains (ε,γ ) is given by,

,xx xx xx xx c c zz zz

core coretop botv v v vU dv dv dv dvσ ε σ ε τ γ σ ε= + + +∫ ∫ ∫ ∫ (6.10.1)

where, topv , botv , and corev are the volume of the top, bottom and core layer, respectively. One may note that, as

the core is taken to be flexible, deformation takes place in the transverse direction ( z direction) and the last term

of equation (1) takes care of that effect. The kinetic energy T can be given by

(b) (c)

(a)

CORE

NA

NA

Lower Skin

Upper Skin



( ) ( ) ( )2 2 2 2 2 2

0 0

1 2L L

t t t b b b c c c c

core corev vT m u w dx m u w dx u dv w dvρ ρ

= + + + + + ∫ ∫ ∫ ∫ . (6.10.2)

Here , and t bm m are the mass per unit length of the top and bottom layer, respectively and cρ is the density of

the core material; tu and bu are the displacement at the neutral axis of the top and bottom layer along x

(longitudinal) direction, respectively; tw and bw are the displacement at top and bottom layer along z (vertical)

direction, respectively (Fig. 6.10.2(c)). Also, cu and cw are the displacement of the core along x and z

directions, respectively and can be given by [1]

( ) , ,,

( / 2) ( / 2)( / 2) b b b x t t t x

c t t t x

u d w u d w zu u d w

c

+ − −= − + , ( )b t

c t

w w zw w

c−

= + (6.10.3)

Here (),x represents the differentiation with respect to x and subscripts t , b, c represent top, bottom and core

layer, respectively.

The non-conservative work done due to the applied load can be given by

( ) 2 2, ,

0 0

1 2L L

nc t x b xW Pw dx Pw dx

= + ∫ ∫ . (6.10.4)

The following non-dimensional parameters are used in this analysis. 2

0 0 (2 )q qP P L E I= , 21 1 ( 2 )q qP PL E I= , 2∗=c c cG A L Eξ , 2

t t tE A L Eφ = , 2 ,=b b bE A L Eφ 2c c cE A L Eφ = ,

( )2 2 = / ( / )( / ) ( / )( / )c t t t b b bg G E c d L d E c d L d+ , 0/ ,t t t= /x x L= , /q qu u L= , /q qw w L= ,

/q qm m m= , /c cm m m= . (6.10.5)

Here, 0P and 1P are, respectively, the non-dimensional static and dynamic load amplitudes; , andq q qE I A are

the Young’s modulus, moment of inertia and the area of cross-section of the qth layer (q equal to t for top layer

and b for bottom layer); t t b bE E I E I= + ; cE , cA and cm are the Young’s modulus, area of cross-section and

mass per unit length of the core, respectively. The non-dimensional time, 4 (1/ 2)0 ( / )t mL E= , where m is the total

mass per unit length. The complex shear modulus of the viscoelastic core is given by * (1 )c c cG G jη= + , where



cG is the phase shear modulus, 1j = − and cη is the core loss factor. The non-dimensional term g is known

as the shear parameter of the system.

Using equations (6.10.1-6.10.5), the governing non-dimensional equations of motion and the boundary

conditions are derived by applying the extended Hamilton’s principle. These resulting governing equations of

motion are as follows.

2 2 4, ,

2 4, ,

( / 3) ( /12)( / ) ( / ) ( / 576)( / )1 ( / )( / ) ( / )

( / 24)( / )( / )( / ) ( / 576)( / )1 ( / )( / ) ( / )

( / 6) ( / 6)( / )( / )

t c t c t t xx c t t c c t xxxx

c t b b xx c b t c c b xxxx

c b c t

m m w m d c c L w m d c d c c L w

m d c d c c L w m d c d c c L w

m w m d c c L u

ξ φ

ξ φ

+ − + +

+ + +

+ +

3, ,

3, ,

2 2 2, ,

(1/ 48)( / 6)1 ( / )( / ) ( / )

( /12)( / )( / ) (1/ 48)( / 6)1 ( / )( / ) ( / )

( / ) ( / 4)1 ( / ) ( / ) ( / 4)1 ( / )1 ( / )

t x t c t c c t xxx

c t b x b c t c c b xxx

c t c t t xx c b c t b b xx

m m d c c L u

m d c c L u m m d c c L u

L c w d c w L c w d c d c w

ξ φ

ξ φ

φ ξ φ ξ

− + +

+ + + +

− − + − − + +

+

3, ,

3, ,

2 2 2, ,

( / 2)( / )1 ( / ) ( / )1 ( / )( / 48)( / )

( / 2)( / )1 ( / ) ( / )1 ( / )( / 48)( / )

( /12)( / ) ( / ) ( / ) 0

c t t x c c t t t xxxxx

c t b x c c t b b xxxxx

t t t xxxx t xx

L c d c u d c c L u

L c d c u d c c L u

d c c L w PL E w

ξ ξ φ φ

ξ ξ φ φ

φ

+ + +

− + − +

+ + =

(6.10.6)

2 4, ,

2 2 4, ,

( / 6) ( / 24)( / )( / )( / ) ( / 576)( / )1 ( / )( / ) ( / )

( / 3) ( /12)( / ) ( / ) ( / 576)( / )1 ( / )( / ) ( / )

( /12)( / )( / )

c t c t b t xx c t b c c t xxxx

b c b c t b xx c b b c c b xxxx

c b

m w m d c d c c L w m d c d c c L w

m m w m d c c L w m d c d c c L w

m d c c L

ξ φ

ξ φ

+ + +

+ + − + +

−

3, , ,

3 2, ,

2 2,

(1/ 48)( / 6)1 ( / )( / ) ( / ) ( / 6)( / )( / )

(1/ 48)( / 6)1 ( / )( / ) ( / ) ( / ) ( / 4)1 ( / )1 ( / )

( / ) ( / 4)1 ( / )

t x t c b c c t xxx c b b x

b c b c c b xxx c b c t b t xx

c t c b b xx

u m m d c c L u m d c c L u

m m d c c L u L c w d c d c w

L c w d c w

ξ φ

ξ φ φ ξ

φ ξ

− + + −

+ + + + − + +

− − + +

,

3, ,

3 2 2 2, , ,

( / 2)( / )1 ( / )

( / )1 ( / )( / 48)( / ) ( / 2)( / )1 ( / )

( / )1 ( / )( / 48)( / ) ( /12)( / ) ( / ) ( / ) 0

c b t x

c c b t t xxxxx c b b x

c c b b b xxxxx b b b xxxx b xx

L c d c u

d c c L u L c d c u

d c c L u d c c L w PL E w

ξ

ξ φ φ ξ

ξ φ φ φ

+

+ + − +

− + + + =

(6.10.7)

3, ,

3, ,

2,

2

( / 6)( / )( / ) ( / 288)( / )( / ) ( / )

( /12)( / )( / ) ( / 288)( / )( / ) ( / )

(1/ 24)( / 6)( / ) ( / ) ( / 3)

(1/ 24)( / 6)( / ) ( / )

c t t x c t c c t xxx

c b b x c b c c b xxx

t c c c t xx t c t

b c c c

m d c c L w m d c c L w


m m c L u m m u

m m c L

ξ φ

ξ φ

ξ φ

ξ φ

−

− −

+ + − +

− +

, ,

2 2, , ,

2 2,

( / 6) ( / 2)1 ( / )( / )

( / 2)( / )1 ( / ) ( / ) ( / )( / 24)( / )

( / ) ( / )( / 24)( / ) 0

b xx c b c t t x

c b b x t t xx c t c c t t xxxx

c b c c b b xxxx

u m u d c L c w

L c d c w u L c u c L u

L c u c L u

ξ

ξ φ ξ ξ φ φ

ξ ξ φ φ

− + +

+ + + − −

+ + =

(6.10.8)



3

, ,

3, ,

2,

( /12)( / )( / ) ( / 288)( / )( / ) ( / )

( / 6)( / )( / ) ( / 288)( / )( / ) ( / )

( / 6) (1/ 24)( / 6)( / ) ( / ) ( / 3)

(1/ 24)( / 6)( /

c t t x c t c c t xxx

c b b x c b c c b xxx

c t t c c c t xx b c b

b c



m u m m c L u m m u

m m c

ξ φ

ξ φ

ξ φ

+

− +

− − + − +

+ +

2, ,

2 2, ,

2 2, ,

) ( / ) ( / 2)1 ( / )( / )

( / 2)( / )1 ( / ) ( / ) ( / )( / 24)( / )

( / ) ( / )( / 24)( / ) 0

c c b xx c t t x

c b b x c t c c t t xxxx

b b xx c b c c b b xxxx

L u d c L c w

L c d c w L c u c L u

u L c u c L u

ξ φ ξ

ξ ξ ξ φ φ

φ ξ ξ φ φ

− +

− + + +

+ − − =

(6.10.9)

As the above equations of motion (6.10.6-6.10.9) are in space and time co-ordinates, generalized

Galerkin’s principle is used to reduce these equations to their temporal form. For multi-mode

discretization one may take 2 3

4

1 1 2 1

3 1

( ) ( ), ( ) ( ), ( ) ( ) and

( ) ( ).

N N N

t p p b q q t r r

N

b s s

p q N r N

s N

w f t w x w f t w x u f t u x

u f t u x

= = + = +

= +

= = =

=

∑ ∑ ∑

∑ (6.10.10)

Here, N is a positive integer representing the number of modes taken in the analysis, and ( )pf t , ( )tf q

, ( )tf r and ( )tf s are the generalized co-ordinates and ( )pw x , ( )xwq , ( )xur and ( )xus are the shape

functions chosen to satisfy as many as the boundary conditions. The resulting equation of motion

becomes

[ ] [ ] [ ] 1 cosM f K f P t H fω φ+ − = . (6.10.11)

Here, ( ) ( ) tdd=⋅ , ,TT T T T

p q r sf f f f f= and [ ] [ ] [ ]1 0= −K K P H ,

where

[ ]

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

M M M MM M M M

MM M M MM M M M

=

, [ ]

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

11 12 13 14

21 22 23 241

31 32 33 34

41 42 43 44

K K K KK K K K

KK K K KK K K K

=



[ ]

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

11

22

HH

H

φ φ φφ φ φφ φ φ φφ φ φ φ

=

, φ and [ ]φ are null matrices.

The elements of the various sub matrices are given below.

( ) ( ) ( )( )

( )( )( )( ) ( )

1 12

110 0

14

0

3 12

576 1 / /

t c i j c t i jij

c t t c c i j

M m m w w dx m d L w w dx

m d c d c c L w w dxξ φ

′ ′= + +

′′ ′′+ +

∫ ∫

∫

( ) ( ) ( )( )

( )( )( )( ) ( )

1 12

120 0

14

0

6 24

576 1

c i j c t b i jij

c b t c c i j

M m w w dx m d d l w w dx


′ ′= −

′′ ′′+ +

∫ ∫

∫

( ) ( )( )( )( ) ( ) ( )( ) 1 1

313

0 0

6 1 48 1 6t c t c c i j c t i jijM m m d c c L w u dx m d L w u dxξ φ

′′ ′ ′= + + +

∫ ∫

( ) ( )( )( ) ( ) ( )( ) 1 1

314

0 0

(1 48) 6 1 /12b c t c c i j c i jijM m m d c L w w dx m L w u dxξ φ

′′ ′ ′= − + + +

∫ ∫

( ) ( ) ( )( )

( )( )( )( ) ( )

1 12

210 0

14

0

6 24

576 1

c i j c t b i jij

c t b c c i j

M m w w dx m d d L w w dx


′ ′= −

′′ ′′+ +

∫ ∫

∫

( ) ( ) ( )( )

( )( )( )( ) ( )

1 12

220 0

14

0

3 12

576 1 / /

b c i j c i jij

c b b c c i j

M m m w w dx m c L w w dx


′ ′= + +

′′ ′′+ +

∫ ∫

∫

( ) ( )( )( ) ( ) ( )( ) 1 1

323

0 0

(1 48) 6 1 12 t c b c c i j c i jijM m m d c c L w u dx m c L w u dxξ φ

′′ ′ ′= + + −

∫ ∫



( ) ( )( )( ) ( ) ( )( )1 1

324

0 0

(1 48) 6 1 / / 6 /b c b c c i j c b i jijM m m d c c L w u dx m d L w u dxξ φ

′′ ′ ′= − + + −

∫ ∫

( ) ( )( )( ) ( ) ( )( ) 1 1

331

0 0

288 + 6 /c t c c i j c t i jijM m d c c L u w dx m d L u w dxξ φ

′′ ′ ′=

∫ ∫

( ) ( )( ) ( )( )( ) ( ) 1 1

332

0 0

12 288 /c b i j c b c c i jijM m d L u w dx m d c c L u w dxξ φ

′ ′′ ′= −

∫ ∫

( ) ( )( ) ( ) ( )1 1

233

0 0

(1 24) 6 3t c c c i j t c i jijM m m c L u u dx m m u u dxξ φ

′ ′= − + − +

∫ ∫

( ) ( ) ( )( ) ( ) 1 1

234

0 0

6 (1 24) 6 /c i j b c c c i jijM m u u dx m m c L u u dxξ φ

′ ′= − + +

∫ ∫

( ) ( )( ) ( )( )( ) ( ) 1 1

341

0 0

12 288c t i j c t c c i jijM m d L u w dx m d c c L u w dxξ φ

′ ′′ ′= − +

∫ ∫

( ) ( )( ) ( )( )( ) ( ) 1 1

342

0 0

6 / 288c b i j c b c c i jijM m d L u w dx m d c c L u w dxξ φ

′ ′= +

∫ ∫

( ) ( ) ( )( ) ( ) 1 1

243

0 0

6 (1 24) 6 /c i j t c c c i jijM m u u dx m m c L u u dxξ φ

′ ′= − + +

∫ ∫

( ) ( ) ( )( ) ( ) 1 1

244

0 0

3 (1 24) 6 /b c i j b c c c i jijM m m u u dx m m c L u u dxξ φ

′ ′= − + − +

∫ ∫

( ) ( )( ) ( )( ) ( ) 1 1 1

2 2 211

0 0 0

1 4 1 1 12 /t c i j t c i j c i jijK d c w w dx d L w w dx L c w w dxξ φ φ

′ ′ ′′ ′′= + + +

∫ ∫ ∫

( ) ( ) ( )( )( )

( )( ) ( )( )

1 12

120 0

13

0

1 4 1 1

1 48 1

c i j t b c i jij

t c c b i j

K L c w w dx d c d c w w dx

c L d c w u dx

φ ξ

ξ φ φ

′ ′= − + + +

′′′ ′′− +

∫ ∫

∫

( ) ( )( )( ) ( )( ) ( )( ) 1 1

313

0 0

1 2 / 1 1 48 1t c i j t c c t i jijK L c d c w u dx c L d c w u dxξ ξ φ φ

′ ′′′ ′′= − + − +

∫ ∫

( ) ( )( )( ) ( )( ) ( )( ) 1 1

314

0 0

1 2 / 1 1 48 1t c i j t c c b i jijK L c d c w u dx c L d c w u dxξ ξ φ φ

′ ′′′ ′′= + + +

∫ ∫

( ) ( ) ( )( )( ) 1 1

221

0 0

1 4 1 1c i j t b c i jijK L c w w dx d c d c w w dxφ ξ

′ ′= − + + +

∫ ∫



( ) ( )( ) ( )( ) ( ) 1 1 1

2 2 222

0 0 0

1 4 1 1 12b c i j b c i j c i jijK d c w w dx d L w w dx L c w w dxξ φ φ

′ ′ ′′ ′′= + + +

∫ ∫ ∫

( ) ( )( )( ) ( )( ) ( )( ) 1 1

323

0 0

1 2 1 1 48 / 1b c i j b c c t i jijK L c d c w u dx c L d c w u dxξ ξ φ φ

′ ′′′ ′′= − + − +

∫ ∫

( ) ( )( )( ) ( )( ) ( )( ) 1 1

324

0 0

1 2 / 1 1 48 1b c i j b c c b i jijK L c d c w u dx c L d c w u dxξ ξ φ φ

′ ′′′ ′′= + + +

∫ ∫

( ) ( )( )( ) 1

310

1 2 1 t c i jijK L c d c u w dxξ

′= − +

∫

( ) ( )( )( ) 1

320

1 2 1 b c i jijK L c d c u w dxξ

′= − +

∫

( ) ( ) ( ) ( )( ) ( ) 1 1 1

2 233

0 0 0

/ 1 24t i j c i j c c t i jijK u u dx L c u u dx c L u u dxφ ξ ξ φ φ

′ ′ ′′ ′′= − − −

∫ ∫ ∫

( ) ( ) ( )( ) ( ) 1 1

2 234

0 0

1 24c i j c c b i jijK L c u u dx c L u u dxξ ξ φ φ

′′ ′′= +

∫ ∫

( ) ( )( )( ) 1

410

1 2 / 1 t c i jijK L c d c u w dxξ

′= +

∫

( ) ( )( )( ) 1

420

1 2 1 b c i jijK L c d c u w dxξ

′= +

∫

( ) ( ) ( )( ) ( ) 1 1

2 243

0 0

/ 1 24c i j c c t i jijK L c u u dx c L u u dxξ ξ φ φ

′′ ′′= +

∫ ∫

( ) ( ) ( ) ( )( ) ( ) 1 1 1

2 244

0 0 0

/ 1 24b i j c i j c c b i jijK u u dx L c u u dx c L u u dxφ ξ ξ φ φ

′ ′ ′′ ′′= − − −

∫ ∫ ∫

( )1

110

i jF w w dx′ ′= ∫

( )1

220

i jF w w dx′ ′= ∫



In the above equations () () x′ = ∂ ∂ .

Equation (6.10.11) is a set of coupled Mathieu-Hill equations with complex coefficients. For numerical

calculations following shape functions (Ray and Kar [7]) are considered.

For simply supported beam

( ) ( )sinpw x p xπ= , ( ) ( )xqxwq πsin= , ( ) ( )xrxur πcos= , and ( ) ( )xsxus πcos= . (6.10.12)

These shape functions satisfy all the boundary conditions. Here 1,2 ,p N= ,

( ) ,q q N= − ( )2r r N= − and ( )3s s N= − . (6. 10.13a)

For clamped-free beam, the shape functions are as follows (Ray and Kar [7])

( 1)2 2 1 2

2 ( 3)2 1 2

( 1)

( ) ( 3)( 2) ( 2)( 1) 2( 3)( 1) ( 2) ( 1) ( 2)( 1)( 2)( 1) ( 1) ( 1) / ( 3)( 2)( 1) ( 3) ,

( ) ( 1)

ii

i

k kk

w x i i i i x i i i i i i i ii i i i i i i i i i x

u x k x kx

+

+

+

= + + + + −µ + + + µ − + µ + + + −µ + + + −µ + + −µ + + + + − + µ

= + −(6.10.13b)

Here and i k are same as the previous boundary conditions.

If [ ] L is a normalized modal matrix of [ ] [ ],1 KM − then the linear transformation

[ ] f L U= , (6.10.14)

transforms equation (6.10.11) to,

4

2

1( ) 2 cos 0

N

q q q qp pp

U U t b Uω ε ω∗ ∗

=+ + =∑ , =q 1….4N; (6.10.15)

where ( )2∗qω are the distinct eigen values of [ ] [ ]KM 1− and ∗

qpb are the elements of

[ ] [ ] [ ] [ ][ ]1 1B L M H L− −= − . Also, <= 21Pε 1 for the present analysis. The complex frequency and forcing

parameters in terms of real and imaginary parts are given by



, ,q q R q Ijω ω ω∗ = + and , ,qp qp R qp Ib b jb∗ = + . (6.10.16)

The boundaries of the regions of instability for simple and combination resonances are obtained by the modified

Hsu’s [9] method. When the system is excited at a frequency nearly equal to twice the natural frequencies

principal parametric resonance and when it is excited near a frequency, which is equal to the sum or differences

of any two modal frequencies combination resonances of sum or difference types take place. Following relations

are used to obtain the boundaries of the regions of instability for simple and combination resonances [7].

(1) Simple resonance case

,( / 2) Rαω ω− < 14 αχ , α = 1, 2…..4N (6.10.17)

where αχ =2 2 2

, , 2,2

,

4 (16R I

IR

b bαα ααα

α

εω

ω

+−

. (6.10.18)

(2) Combination resonance of sum type

, ,( )R Rα βω ω ω− + < αβχ (6.10.19)

when damping is present,

αβχ =

1/ 22, , , , , ,

, ,1/ 2, ,, ,

( ) 4 ( )16 ,

4( )I I R R I I

I IR RI I

b b b bα β αβ βα αβ βαα β

α βα β

ω ω εω ω

ω ωω ω

+ + −

(6.10.20)

and for the undamped case,

1/ 2

, ,

, ,

.

R R

R R

b bαβ βααβ

α βχ ε

ω ω

=

α β≠ , ,α β = 1, 2…..4N (6.10.21)

(3) Combination resonance of difference type

, ,( )R Rβ α αβω ω ω− − < Λ α β> , ,α β = 1, 2…..4N (6.10.22)

when damping is present,



1/ 22, , , , , ,

, ,1/ 2, ,, ,

( ) 4 ( )16

4( )I I I I R R

I IR RI I

b b b bα β αβ βα αβ βααβ α β

α βα β

ω ω εω ω

ω ωω ω

+ − Λ = −

, (6.10.23)

and for the undamped case,

1/ 2, ,

, ,

.

R R

R R

b bαβ βααβ

α βε

ω ω

Λ = −

(6.10.24)

Numerical Results and Discussions

Here the parametric instability regions of a three-layered symmetric sandwich beam with simply

supported, and clamped-free boundary conditions have been determined numerically using MATLAB.

For visco-elastic materials, core loss factor ( cη ) is a measure energy dissipation capacity and the shear

parameter ( )2 = / 2 ( / )( / )c t t tg G E c d L d is a measure of stiffness of the material and is important in

determining how much energy gets into the visco-elastic material. So these two parameters are varied

in determining the instability regions for the parametrically excited beams. Also the effects of core and

skin thickness on the instability regions are studied for all these boundary conditions. In the parametric

instability regions shown in the following figures, the regions enclosed by the curves are unstable and

the regions outside the curves are stable. Here the ordinate 1P is the amplitude of non-dimensional

dynamic load and the abscissaϖ is the non-dimensional forcing frequency. Following physical

parameters are taken for the numerical analysis. The span of the beam, L=300 cm, width, b=50 mm, the

top and bottom face thickness 2== bt dd mm and the core thickness, c = 30 mm. The non-

dimensional static load amplitude 0 0.1P = for all the figures except it is specifically mentioned. The

top and bottom faces are of steel and the core is of soft plastic foam (Divinycell H60). The mechanical

properties of steel and Divinycell H60 are given in Table (1).

Table 1:Material properties of sandwich beam [10]

Material

Young’s modulus

E, Gpa

Shear modulus

G, Gpa

Poisson’s ratio

ν

Density ρ, kg/m3

Steel

210

81

0.3

7900



Divinycell H60

0.056

0.022

0.27

60

Simply supported beam

Using the shape functions given in equation (6.10. 12) the instability regions for the simply supported

beam are determined and shown in Figures (6.10.3-6.11.6) for the first three modes. Figure 6.10.3

shows the parametric instability regions obtained using both the higher-order theory and classical

theory [10] for simple resonances. One may observe that for all the three modes, the region of

instability starts at a lower frequency for higher order theory in comparison to the classical theory,

which is due to the fact that, the core is considered to be more flexible in higher order theory than in

case of classical theory. Also, it is clearly observed from these figures that the instability region is

wider in case of higher order theory as compared to the classical theory. With change in 0P (say 0P

=0.1), while instability region with higher order theory remains almost unchanged, it is observed that

for lower value of 1P , the instability regions with classical theory shifts towards left.

Figure 6.10.3: Comparison of instability regions using higher-order and classical theories, 0 0.5, cP η= = 0.1;

g=0.05; , higher-order theory; ------, classical theory.



Figure 6.10.4: Effect of shear parameter on instability regions for =cη 0.0. , g = 0.05; -----, g = 0.1; +++

, g = 0.5.

Figure 6.10.5: Effect of shear parameter on instability regions for =cη 0.3. , g = 0.05; -----, g = 0.1; +++,

g = 0.5.

Figures (6.10.4-5) show the influence of core loss factor ( cη ) and the shear parameter (g) upon the instability

region obtained by using higher order theory. It is clearly observed that increase in core loss factor improves the

stability by shifting the instability zones upwards and reducing the area of instability, which is similar to those,

obtained by classical theory. It is also observed that with increase in shear parameter stability of the system

improves. From the above figures it is clearly understood that to get a more stable system one may go for higher

value of core loss factor ( cη ) and shear parameter (g).

Clamped-Free beam.

Using the shape functions (equation (6.10.15)) for the clamped-free beam, the instability regions for the first

three modes are determined and shown in Figures (6.10.6-8).



Figure 6.10.6: Comparison of instability regions using higher order and classical theories for 0 0.5, cP η= = 0.1,

g = 0.05. , higher-order theory; ------, classical theory.

Using higher order theory and classical theory the parametric instability regions for simple resonances

are shown in Figure 6.10.6. Here also, higher order theory gives a conservative design for lower modes.

Figures (6.10.7-8) show the influence of core loss factor ( cη ) and the shear parameter (g) upon the instability

region and it is observed that with increase in core loss factor and shear parameter stability of the system

improves.

Figure 6-10.7: Effect of shear parameter on instability regions for =cη 0. , g = 0.05;

-----, g = 0.1; +++, g = 0.5.



Figure 6.10.8: Effect of shear parameter on instability regions for =cη 0.3. , g = 0.05; -----, g = 0.1; +++, g =

0.5.

For all the boundary conditions the system is always found to be stable at combination resonances of sum and

difference type. In these cases, for simple resonances it is observed that with increase in shear parameter the

instability plot moves upward implying that there exists critical forcing amplitude below which the system is

always stable. For example, when a cantilevered sandwich beam with cη =0.3 and g= 0.1 is excited near twice the

first natural frequency ( 8.2ω ), the system will not vibrate if the forcing amplitude is less than 0.485 (point Pc

on figure 6.10.8). But for the same cη and g = 0.05, with same amplitude of forcing, the system will vibrate at a

slightly less frequency (say 7.8ω ). Again with increase in shear parameter, the instability region shifts

towards right and hence, for same forcing amplitude, the system becomes unstable at a higher frequency. As the

shear parameter ( )2 = / 2 ( / )( / )c t t tg G E c d L d , is a function of dimension and material properties of both skin

and core material, using the above stability charts, a designer will be able to construct sandwich beams having

very less or vibration free structures.

REFERENCES

1. FROSTIG, Y., and BARUCH, M., Bending of sandwich beams with transversely flexible core.

American Institute of Aeronautics and Astronautics Journal, 27, 523-531, 1990.

2. FROSTIG, Y., and BARUCH, M., Free vibrations of sandwich beams with a transversely flexible core:

A higher order approach. Journal of Sound and Vibration, 176(2), 195-208, 1994.

3. FROSTIG, Y., Buckling of sandwich panels with flexible core-high order theory. International Journal

of Solids and Structures, 35(3-4), 183-204,1998.

Pc



4. FROSTIG, Y., and THOMSEN, O.T., High-order free vibration of sandwich panels with a flexible core.

International Journal of Solids and Structures, 41, 1697-1724, 2004.

5. SAITO, H., and OTOMI, K., Parametric response of viscoelastically supported beams. Journal of Sound

and Vibration, 63, 169-178, 1979.

6. KAR, R.C., and SUJATA, T., Dynamic stability of a tapered symmetric sandwich beam. Computers &

Structures, 40, 1441-1449, 1991.

7. RAY, K., and KAR, R.C., Parametric instability of a sandwich beam with various boundary conditions.

Computers & Structures, 55, 857-870, 1995.

8. RAY, K., and KAR, R.C., Parametric instability of multi-layered sandwich beams. Journal of Sound

and Vibration, 193(3), 631-644, 1996.

9. HSU, C.S., On the parametric excitation of a dynamic system having multiple degrees of freedom.

Journal of Applied Mechanics, ASME, 30, 367-372, 1963.

10. Dwivedy S. K., Sahu K. C. and Babu Sk., Parametric instability regions of three layered soft-cored

sandwich beam using higher order theory, Journal of Sound and Vibration, 304, 326-344, 2007.

Module 6 Lecture 11

Parametrically excited continuous system

Case study: Nonlinear Vibration of a Magneto-Elastic Cantilever Beam With Tip Mass

In this work the effect of the application of alternating magnetic field on the large transverse vibration

of a cantilever beam with tip mass is investigated. The governing equation of motion is derived using

the D’ Alembert’s principle which is reduced to its non-dimensional temporal form by using the

generalize Galerkin’s method. The temporal equation of motion of the system contains the



nonlinearities of geometric and inertial type along with parametric excitation and non-linear damping

term. Method of multiple scales is used to determine the instability region and frequency response

curves of the system. The influences of the damping, tip mass, amplitude of magnetic field strength,

permeability and conductivity of the beam material on the frequency response curves are investigated.

These perturbation results are found to be in good agreement with those obtained by numerically

solving the temporal equation of motion and experimental results.

Fig. 6.11.1: Schematic diagram of a flexible single-link cantilever beam with tip mass.

Figure 6.11.1 shows a flexible cantilever beam with a tip mass M. The beam is subjected to a harmonic

transverse magnetic field 0 cosmB B t= Ω where mB and Ω are respectively, the amplitude and

frequency of the magnetic field strength. In this work, the flexible cantilever beam with tip mass is

modeled as an Euler-Bernoulli beam with a tip mass. For the purpose of completeness a brief derivation

s

v

u M

X

Y



of the equation of motion using d’ Alembert’s principle is given below. The bending moment ( )M s of

the beam at a distance s from the fixed end (Fig. 1) can be expressed as [7,11]

( ) 212ss s ssM s E I v v v ≈ +

. (6.11.1)

Here, v is the transverse displacement of the beam. ( )s is the first derivative with respect to s. One

may write the inextensibility condition of the beam in terms of longitudinal displacement ( )tu ,ξ and

transverse displacement ( )tv , ξ as [7]

( )22 1 1s sv u+ + = . or, ( ) ( )1

2 2, 10

u t v dη

ξξ = ξ − − η∫ . (6.11.2)

Here ,ξ η are the integration variables. Considering the inertia forces , , and Au Av M u M vρ ρ , and

using the d’ Alembert’s principle, one may write Eq. (6.11.1) as follows

( ) ( ) ( ) 0LM s M s M sξ− − = . (6.11.3)

Here ( )M sξ is the summation of the moment due to inertia force of beam and the moment due to

external magnetic force a distance ξ from the roller support and the couple due to magnetic field.

( )LM s is the moment due to inertia force for the pay load at the tip of the manipulator. The

expressions for these moments are given below.

( ) ( )sin cos

sin cos ,

d

L LM s Au d d Av C d d

s s s sL

p d d c ds s s

vξ ξ

= − ρ θ η ξ − ρ + θ η ξ∫ ∫ ∫ ∫ξ

ξ ξ− ξ θ η − θ η∫ ∫ ∫

(6.11.4)



Here, p and c are the body force and body couple of the beam due to the magnetic field 0B which are

expressed as [3, 5, 6, 9]

2 2 2 20 0

0

1 11 , and 2 20

s s s s s sr

mp h dB v v v v v v d c h dB vs = − − − =∫

ξ χσ ξ

µ µ. (6.11.5)

Also, ( ) sin cosL L

M s M u d M v dL s s= − θ ξ − θ ξ∫ ∫ . (6.11.6)

By differentiating Eq. (6.11.3) twice with respect to s and applying the Leibnitz’s rules one may obtain

the following governing differential equation of motion.

( ) ( )2 3 2ξ

1 32

s

ssss s ssss s ss sss ss s s ss0

v v v dξ ξEI v v v v v v v ρAv v v + ξ

+ + + ∫+ +

( )( ) ( ) ( )2 2LL

d s ss ss ξ ξ ξ ξ ξ ξs s 0 0

dρ Av C v M v v v d dη M dv ρA v v v v v vξ ξ

η+ + − ξ + ξ∫

+ + ∫ ∫ ∫

( ) ( ) ( ) ( )( )2 2 21 1 11 1 1 02 22

L

ss s s s ss ss ds

dcv ρAv C v pd pv v v v v cds

v −− + − ξ − − + + =∫

. (6.11.7)

To obtain the temporal equation of motion, one may discretize the governing equation of motion

(6.11.7) by using following assumed mode expression.

( ) ( ) ( ),v s t r s q t= ψ . (6.11.8)

Here, r is the scaling factor; q(t) is the time modulation and ( )sψ is the eigen-function of the cantilever

beam with tip mass, which is given by [7]

( ) ( )sin sinh( ) cos cosh sin sinhcos cosh

L Ls s s s sL L

β + βψ = − β − β + β − β β + β

. (6.11.9)



One may determine Lβ from the following equation.

( )1 cos cosh cos sinh sin cosh 0L L m L L L L L+ + − =β β β β β β β . (6.11.10)

Following non-dimensional parameters are used in this analysis.

Lsx = , e t=τ ω ,

e

Ω=ωω

, rrL

= , MmAL

=ρ

, 4ALEIρ

χ = . (6.11.11)

Substituting Eq. (6.11.9) into Eq. (6.11.6) and using the generalized Galerkin’s method, one may obtain

the resulting non-dimensional temporal equation of motion, which can be expressed as

( ) ( )( )( )

3 2 21 2 3 1

21

2 cos 2

- 1 cos 2 0.

q q q q q q q q f q

k qq

+ εζ + + ε α + α + α − ε ωτ

ε + ωτ =

. (6.11.12)

The expressions for the coefficients ( 1 2 3 1 1i.e. , , , , ,f kζ α α α ) in this equation are given below.

The natural frequency of the lateral vibration of an elastic beam

2

151 14 2

14 0 14

22

mm

r

Be

h d hE I hmL h mL h

χω = − µ µ

,

( )22

151 14

14 0 1 1

22

1 1mmm

r

BL

h d L hE I h BmL h E I h

χ= − = ω − µ µ

. (A1)

Here, 1 14

14

2L

E I hmL h

ω = , 2

15

0 1 1

2and .

2m

r

mm

B h d L hBE I h

χ = µ µ

Damping ratio due to the viscous damping to the system,



2d L

e e

Cm

ωζ = = µ = µδ ε ω ω

, (A2)

Coefficient of the nonlinear geometric term 3q =

32 41

14 14 14

2

4 2 32

e

hh hEI rh h hm L

α = + +

εω , (A3)

Coefficient of the nonlinear inertia term 2q q=

5 6 7 8 9 102

14 14 14 14 14 14

2 h h h h h hr m mh h h h h h

α = + + − − − ε

, (A4)

Coefficient of the nonlinear inertia term 2q q =

1311 123

14 14 14

2 hh hr mh h h

α = − − ε

, (A5)

Coefficient of the parametric excitation termcos(2 ) qωτ =

01 2

2

2 2r

e c

B ffB

= = δω , where

22

2m

rBB = and

20 1 1

14

2 r

mc

E IL hBh d h

µ µ= χ

. (A6)

Coefficient of the nonlinear damping terms ( ) 21 cos(2 ) q q+ ωτ =

22 16 17

114 142

m

e

B h d h hk rm h h

σ= − − + ω

. (A7)

Here,



( ) ( )41

10

d xh x dx

dxψ

= ψ∫ , ( ) ( )

321

20

d xh x dx

dx ψ

= ψ ∫ ,

( ) ( ) ( )41

30

d x d xh x dx

dx dxψ ψ

= ψ∫

( ) ( ) ( ) ( )2 31

4 2 30

d x d x d xh x dx

dx dx dxψ ψ ψ

= ψ∫ , ( ) ( ) 2

1

50 0

( )x dd x

h d x dxdx d

ψ ξψ = ξψ ξ

∫ ∫ ,

( ) ( ) ( ) ( )21 1

6 20 x

d x d xh d x dx

dx dxψ ψ

= ψ ξ ξψ∫ ∫ , ( ) ( ) ( )

212

7 20

( )d x d x

h x dxdx dxψ ψ

= ψ∫ ,

( ) ( ) ( )221 1

8 20 0x

d x d xh d d x dx

dx d

ηψ ψ = ξ ηψ ξ ∫ ∫ ∫ ,

( ) ( ) ( )2

21

9 20 0

x dd xh d x dx

dx d

ψ ξψ = ξψ ξ

∫ ∫ ,

( ) ( )( )21

210

0

d xh x dx

dxψ

= ψ ∫ ,

( ) ( ) ( )2

1

110 0

x dd xh d x dx

dx d

ψ ξψ = ξψ ξ

∫ ∫ ,

( ) ( ) ( )2

21 1

12 20 0x

dd xh d d x dx

dx d

η ψ ξψ = ξ ηψ ξ

∫ ∫ ∫ , ( ) ( ) ( )

221

13 20 0

x dd xh d x dx

dx d

ψ ξψ = ξψ ξ

∫ ∫ ,

( )( )1

214

0

h x dx= ψ∫ , ( ) ( )

21

15 20

d xh x dx

dx ψ

= ψ ∫ ,

1

160 0

( ) ( ) ( )xd x dh d x dx

dx d ψ ψ ξ

= ξ ψ ξ ∫ ∫ ,

and 1 12

17 20 0

( ) ( ) ( )x

d x dh d d x dxdx d

ξ ψ ψ ξ= η ξ ψ ξ ∫ ∫ ∫ .

Here one may observe that the non-dimensional temporal Eq. (6.11.12) has parametric term

( )1 cos 2f qωτ and nonlinear damping term ( )( ) 21 1 cos 2k qq+ ωτ , along with cubic geometric ( 1

3qα )

and inertial ( 2 32 2q q q qα +α ) nonlinear terms. Hence, it may be noted that the temporal equation of



motion Eq. (6.11.12) contains many nonlinear terms and it is very difficult to find the exact solution.

Hence one may go for the approximate solution by using the perturbation method. Here method of

multiple scales is used.

In method of multiples scales, the displacement q can be represented in terms of different time scales

0 1( , )T T and a book keeping parameter ε as follows.

( ) ( ) ( )20 0 1 1 0 1( ; ) , ,q q T T q T T Oτ ε = + ε + ε . (6.11.13)

Here, 0T τ= , and 1T ετ= . The transformation of first and second time derivatives are given by

)( 210 εε

τODD

dd ++= , (6.11.14)

)(2 210

202

2εε

τODDD

dd ++= . (6.11.15)

where, 00

DT∂

=∂

, and 11

DT∂

=∂

. Substituting Eqs. (6.11.13- 6.11.15) into Eq. (6.11.12) and equating the

coefficient of like powers ofε , yields the following equations.

Order :0ε 20 0 0 0D q q+ = , (6.11.16 )

Order :1ε 20 1 1 0 1 0 0 02 2D q q D D q D q+ = − − ζ ( ) ( )23 2 2

1 0 2 0 0 0 3 0 0 0q D q q D q q−α −α −α

( ) ( )( )( ) 21 0 0 1 0 0 0 0cos 2 1 cos 2f T q k T D q q+ ω + + ω . (6.11.17)

General solutions of Eq. (6.11.16) can be written as



0 1 2 0 2 0( , ) exp( ) ( , ) exp( )q A T T iT A T T iT= + − . (6.11.18)

Substituting Eq. (6.11.18) into Eq. (6.11.17) leads to

( )2 20 1 1 0 0 1 2 3 1 02 exp( ) 2 exp( ) 3 3 exp( )D q q i A iT i A iT ik A A iT′+ = − − ζ − α − α +α −

( ) ( ) ( )3 3 11 2 3 1 0+ exp 3 + exp 3 + exp (2 1) exp (2 1)

20 0fA iT ik A iT A i A i T −α +α +α ω− + ω−

3 2 21 1 10 0 0+ exp (2 3) exp (2 1) exp (2 1)

2 2 2ik ik ikA i T A A i T A A i Tω+ + ω+ − ω−

310+ exp (3 2 )

2ik A i T cc− ω + . (6.11.19)

Here, cc is the complex conjugate of the preceding terms One may observe that any solution of Eq.

(6.11.19) will contain secular or small divisor terms when non-dimensional frequency of magnetic field

strength (ω ) is nearly equal to 1 which may be called as simple resonance case. In this case, one may

use detuning parameter σ to express the nearness of ω to 1, as

( )1O and ,1 =+= σσεω . (6.11.20)

Substituting Eq. (6.11.20) into Eq. (6.11.19), one may obtain the following secular or small divisor

terms.

( ) 20 0 1 2 3 1 02 exp( ) 2 exp( ) 3 3 exp( )i A iT i A iT ik A A iT′− − ζ − α − α +α −

( ) ( ) ( )3 21 1 11 1 1exp 2 exp 2 exp 2 0.

2 2 2f k kA T i A T i A A T+ σ + − σ − σ = (21)



Putting A equal to ( ) ( )11

1 2

i Ta T e β and 12 2Tγ = σ − β into Eq. (6.11.21) and separating the real and

imaginary terms, one may a set of reduced equations as given below.

31 1 sinγ8 4k fa a a a= −ζ + + , (6.11.22)

3 33 11 2 1

1 3 12 sin cosγ4 3 4 2

fa a a a k aαω− γ = − α −α + + γ + ε . (6.11.23)

One may observe from the Eqs. (6.11.21)-( 6.11.22) that the system possesses both trivial and

nontrivial responses. Hence one may determine both responses by solving Eqs. (6.11.22, 6.11.23)

simultaneously. To find the stability of the steady state responses, one may perturb the above Eqs.

(6.11.22, 6.11.23), by substituting 1oa a a= + and 0 1γ = γ + γ where 00 γ,a are the singular points, and

then investigate the eigenvalues of the Jacobian matrix (J) which is given by

21 1 10 0 0 0

31 2 0 0 1 0 2 1

0 1 0 0

3 sin cos8 4 4

3 1+ sin 1 cos sin2 3 24 2

k f fa a

J a a k fa k

−ζ + + γ γ

α= − α −α + γ γ − γ

. (6.11.24)

It may be noted that the system will be stable if and only if all the real parts of the eigen-values are

negative.

For trivial state instability region, one may use the following expression for the transition curve for

simple resonance case which has been obtained by finding the eigen-values of Jacobian matrix (J)

given in Eq.( 6.11.24).



( )2

2 20 4 .2 4L

f O Ω ε

Ω = = δ± − µ + ε ω δ (6.11.25)

Here, the expression for 0, , and fδ µ are given in appendix [7].

It may be noted that this simple expression has been obtained by using first order method of multiple

scales is different from the expression given in the work of Pratiher and Dwivedy [7], which was

obtained by using the second order method of multiple scales.

Now the first order non-trivial steady state response of the cantilever beam with a tip mass can be given

by

( )1cos2

q a = ωτ− γ

. (6.11.26)

Here for numerical simulations, a steel beam similar to that considered in the work of Wu [5, 6] with

length L = 0.5 m, width d = 0.005 m, depth h = 0.001 m, Young’s Modulus E = 1110941 ×. N/m2, mass

of the beam per unit length m = 0.03965 kg, and the permeability of the vacuum, 0µ = 61026.1 −× Hm-1

have been considered. Using these parameters, the reduced Eqs. (6.11.22, 6.10.23), have been solved

numerically to obtain the instability regions and the frequency response curves. In the instability plot,

the regions bounded by the curves are unstable and regions outside the curves are stable. In the

frequency response curves dotted and solid lines represent, respectively the unstable and stable

response of the system. The effect of the amplitude of magnetic field strength ( mB ), damping ( dC ), tip

mass (M), material conductivity (σ), and relative permeability of the material ( rµ ) on the frequency

response have been investigated.



Fig. 6.11.2: The region of instability of a cantilever beam with tip mass subjected to magnetic field,

-.-.-.-.-. Moon and Pao’s theoretical result, - - - - - Moon and Pao’s experimental result, and ---- the

present result.

For simple resonance case, the beam is subjected to a transverse magnetic field with a frequency nearly

equal to the natural frequency of the system. Here, the instability regions are plotted in ( )2/ LΩ ω Vs

( )2/r cB B plane similar to the work of Moon and Pao [2], Wu [5, 6], and Pratiher and Dwivedy [7]. The

experimental and theoretical results of Moon and Pao [2] are also being plotted in Fig.6.11.2 for

comparison with the present result. It is found from Fig.6.11.2 that the result obtained in the present

work is in good agreement with the experimental results Moon and Pao [2]. The accuracy of the

instability region obtained by using the first order method of multiple scales can be verified by

numerically solving the temporal equation of motion (6.11.12) and plotting the time response (Fig.

6.11.3) for two different points A and B as marked in Fig.6.11.2. Figure 6.11.3(i) clearly shows that the

response is stable and Fig.6.11. 3(ii) shows that the response is unstable which are in good agreement



with the result shown in Fig.6.11. 2. Hence, one may use the first order closed form solution (Eq.

6.11.25) for finding the instability region instead of going for a second order solution as reported in the

work of Pratiher and Dwivedy [7]. But when more accurate result is required, one may use the

expression given in the work of Pratiher and Dwivedy [7].

Fig.6.113. (a) Time response for the point A and (b) time response for point B marked in Fig.6.11.2.

Figure 6.11.4 shows the frequency response curve for four different values of amplitude of magnetic

field strength mB . From Fig.6.11.4, it may be noted that with increase in mB , though the maximum

response amplitude remain unchanged, the trivial state becomes unstable which is similar to that shown

in Fig.6.11.2. The trivial state becomes unstable by the sub-critical pitchfork bifurcation at 1R , which

ends with a super-critical pitchfork bifurcation at 2R . Here, one may observe that the system has a

tendency to jump up from the unstable trivial state at 1R to the stable non-trivial state at /1R .

Figures 6.11.5(a) and (b) show the transient and steady state response for point C marked in Fig.

6.11.4(c). The solid line and dotted line respectively represent the response of the system with and

without magnetic field. In the presence of magnetic field, it clearly shows that the steady state response

q q

Time (τ) Time (τ)

(a) (b)



has zero response amplitude. Also it may be noted that the free vibration response of the beam shown

as dotted line in Fig. 6.11.5, is reduced by applying the magnetic field.

Fig.6.11.4. Effect of the magnetic field strength ( mB ) on the frequency response curves for 0.02 kgM = ,

20.01 N-s/mdC = , 3000rµ = , 710σ = Vm-1 (a) 0.20mB = Am-1 (b) 0.25mB = Am-1 (c) 0.30mB = Am-1

(d) 0.35mB = Am-1.

(a) (b)

a a

(c)

(d)

a a

A BA

C R2 R1



Fig.6. 11. 5. (a) Transient response and (b) steady state time response for the point C with and without magnetic

field.

The effect of damping dC on the response curves is shown in Fig.6.11.6 and it has been observed that

with increase in dC , while the non-trivial response amplitude remains unchanged the trivial state

unstable region decreases, the sub-critical pitchfork bifurcation point occurs at a higher value of ω and

the corresponding jump length decreases.

Fig.6.11.6: Influence of damping on frequency response curve for 0.02 kgM = , 3000rµ = , 710σ = Vm-1,

0.30mB = Am-1 (a) 20.02 N-s/mdC = ,(b) 20.03 N-s/mdC = .

Similarly one can study the influence of effect of relative permeability ( rµ ), material conductivity (σ)

and mass ratio on the frequency response curves of the system.

(a) (b)

q q

Time (τ) Time (τ)

(a) (b)

a a



Reference

[1] Dwivedy S. K., and Eberhard P., 2006, “Dynamic Analysis of Flexible Manipulators,” Mechanism

and Machine Theory, 41, pp 749-777.

[2] Moon, F.C., and Pao, Y. H., 1969, “Vibration and Dynamic Instability of a Beam-Plate in a

Transverse Magnetic Field,” Journal of Applied Mechanics, 36, pp. 92–100.

[3] Wu, G.Y., Tsai, R., and Shih, Y.S., “The Analysis of Dynamic Stability and Vibration Motions of a

Cantilever Beam with Axial Loads and Transverse Magnetic Fields,” Journal of the Acoustical Society

of ROC, 4, pp. 40–55.

[4] Chen, C.C., and Yah, M.K., 2001, “Parametric Instability of a Beam under Electromagnetic

Excitation,” Journal of Sound and Vibration, 240, pp. 747–764.

[5] Wu, G.Y., 2005, “The Analysis of Dynamic Instability and Vibration Motions of a Pinned Beam

with Transverse Magnetic Fields and Thermal Loads,” Journal of Sound and Vibration, 284, pp. 343–

360.

[6] Wu, G.Y., 2007, “The Analysis of Dynamic Instability on the Large Amplitude Vibrations of a

Beam with Transverse Magnetic Fields and Thermal Loads,” Journal of Sound and Vibration, 302,

pp.167-177.

[7] Pratiher, B., and Dwivedy, S. K., 2007, “Parametric Instability of a Cantilever Beam with Magnetic

Field and Periodic Axial Load,” Journal of Sound and Vibration, 305, pp.904-917.

[8] Kojima, H., and Nagaya, K., 1985, “Nonlinear Forced Vibration of a Beam with a Mass Subjected

to Alternating Electromagnetic Force,” Bull. of the Japan Society of Mechanical Engineers, 28, pp.

468–474.



[9] Lu, Q. S., To, C.W.S., and Huang, K.L., 1995, “Dynamic Stability and Bifurcation of an Alternating

Load and Magnetic Field Excited Magneto-Elastic Beam,” Journal of Sound and Vibration, 181, pp.

873–891.

[10] Nayfeh, A.H., and Mook, D.T., 1995, “Nonlinear Oscillations,” Wiley, New York.

[16] Cuvalci, O., 2000, “The Effect of Detuning Parameters on the Absorption Region for a Coupled

System: a Numerical and Experimental Study,” Journal of Sound and Vibration, 229, pp. 837-857.

Module 6 Lect 12

Parametrically excited System with internal Resonance

In this lecture a case study is taken for a parametrically excited system with internal resonance. The system considered is a uniform cantilever beam of length 𝐿 carriying a mass 𝑚 at an arbitrary position 𝑑 from the fixed end and subjected to base motion ( ) tZtz Ω= cos0 as shown in Fig. 6.12.1.

Fig. 6. 12. 1. Base Excited Cantilever beam with attached mass at arbitrary position

The equation of motion of the beam is given by Kar and Dwivedy[1999] as



( ) ( ) ( )( ) ( )

2 3 2

0

1 13 12 2

0

ssss s ssss s ss sss ss s tt t

L

s sss tt t s stt ss

EI m s d c

m d c d j s d N

ν ν ν ν ν ν ν ν ρ δ ν ν

υ υ ρ δ ξ υ υ ξ δ υ υ

+ + + + − + − +

+ + − + − − − = ∫ (6.12.1)

Subject to the boundary conditions

( ) ( ) ( ) ( ) 0,,0,,0,0,0,0 ==== tLtLtt ssssss υυυυ (6.12.2)

Where

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2

0

2 20

1 12 2

112

s L

s s tttt ttL s s

s

tt stt s s stL

N d d m d d d m z g

sd d L z g J s dL

ξ ξ

ρ υ η ξ δ ξ υ η ξ

δ ξ ξ ρ δ υ υ υ υ

= + − × + −

× − + − − − − +

∫ ∫ ∫ ∫

∫ (6.12.3)

Here, ( ) ( ) ( ) ( )st st ∂

∂=

∂∂

= ,

Here, E, I and 𝜌 are, respectively, the Young’s modulus, the second moment of area of the cross-section of the beam and mass per unit length of the beam; 𝑗0 is the moment of inertia of the concentrated mass 𝑚 about its centroidal axis perpendicular to the X-Y plane; υ is the lateral displacement of the beam; 𝑔, 𝑐 and 𝑧 are, respectively, the acceleration due to gravity, the coefficient of viscous damping and the displacement of the base; and δ is the Dirac delta function. Assuming a solution of Eq. (6.12.1) in the form

( ) ( ) ( ), n nn

s t r s u tυ ϕ∞

=

=∑1

(6.12.4)

Where r is a scaling factor, n sφ is the shape function of the n th mode, and 𝑢𝑛 is the time

modulation of the nth mode. Applying generalized Galerkin’s method and using the following non-dimensional parameters,

,,,,1

1 θθ

ωθτβ nnt

Ld

Lsx ====

12

00 ,,,,θ

φρρ

µλ Ω=====

LrJ

JZZ

TL

mLr

r

(6.12.5)

Eq. (6.12.1) reduces to

2

1 1 1 1

0,2 cosn n nklm k l m klm k l m klm k l m

n n n n n nm mm k l m

u u u u u u u u uu u u f u α β γεξ ω ε φτ ε∞ ∞ ∞ ∞

= = = =

+ + =+ + − +∑ ∑∑∑

(6.12.6)



Here, 1,2,...n = , ( ) ( ) τdd /. = . For details of the coefficients one may refer (Kar and Dwivedy 1999). The small dimensionless parameter ε is the book-keeping parameter to indicate the smallness of damping, non-linearities and excitation. So, we have 𝑛 number of coupled equations with cubic geometric and inertial non-linearities, where 𝑛 represents the number of modes participating in the resulting oscillation. Due to the absence of any internal and external excitation for 3≥n , the amplitude of these higher modes die out in the presence of damping and hence two mode discretization in the Galerkin’s method is sufficient in this particular system.

The approximate solution of Eq. (6.12.6) can be obtained using the method of multiple scales. Let

( ) ( ) ( ) ...,,; 101100 ++= TTuTTuu nnn εετ (6.12.7a)

∞=== ,...,2,1,, 10 nTT εττ (6.12.7b)

Substituting Eqs. (6.12.7a) and (6.12.7b) into Eq. (6.12.6) and equating the coefficients of 0ε and ε to zero, yields

,002

020 =+ nnn uuD ω (6.12.8)

∑∞

=

−+−=+1,

00100012

120 cos22[

mnmnmnnnnnn ufuDDuDuuD φτξω

00200000000000 =+++∑

klmmlk

nklmmlk

nklmmlk

nklm uDuuuDuDuuuu γβα (6.12.9)

Where 00 / TD ∂∂= and 11 / TD ∂∂= . The solution of Eq. (9) is given by

( ) ( ) ccTiTAu nnn += 010 exp ω (6.12.10)

Where cc indicates the complex conjugate of the preceding terms and nA is determined in the following section. Considering the Principal parametric resonance ( )12ωφ ≈ , to express the nearness of φ to

12ω the detuning parameter 1σ is introduced. Also, to account for the internal resonance, the detuning

2σ is used. Hence, we have

,ϕ ω εσ ω ω εσ= + = +1 1 2 1 22 3 (6.12.11)

Substituting Eqs. (6.12.10) and (6.12.11) into Eq. (6.12.9) and eliminating the secular terms, we get for 𝑛 = 1

( ) ( ) ( ) [ ]01221201111'1111 expexp

212 TiAfTiAfAAi σσεεσξω −+−+



( )∑∞

=

=++1

022

121211 0expj

jjje TiAAQAAA εσα (6.12.12)

For n=2,

( ) ( ) 021121'2222 exp

212 TiAfAAi σσεξω −−+

( )∑∞

=

=−++1

023

122112 0expj

jjje TiAAQAAA εσα (6.12.13)

For 3≥n ,

( ) ∑∞

=

=++1

' 02j

njjenjnnnn AAAAAi αξω (6.12.14)

Where a prime denotes the derivative with respect to 𝑇1. Since the higher modes ( )3≥n are neither directly excited by external excitation nor indirectly excited by internal resonance, from Eq. (6.12.14) it can be shown that the response amplitude of these modes die out due to the presence of damping.

Letting ( ) ( ) 11 exp21 TiTaA nn β= (where na and nβ are real) in Eqs. (6.12.12) and (6.12.13) and then

separating into real and imaginary parts, one obtains

( ) ( ) 212121111'1111 sin2sin

212 γγγξω −+−+ afafaa ( ) 03sin25.0 21

21212 =−+ γγaaQ (6.12.15a)

( )

( )

'1 1 1 1 11 1 1 12 2 1 2

22 2

1 1 12 2 1 1 21

1 12 cos 2 cos2 2

1 1 cos 3 04 4e j j

j

a f a f a

a a Q a a

ω γ σ γ γ γ

α γ γ=

− − + − +

+ − =∑ (6.12.15b)

( ) ( ) ( )' 32 2 2 2 21 1 2 1 21 1 2 1

1 12 sin sin 3 02 4

a a f a Q aω ξ γ γ γ γ+ − − + − = (6.12.15c)

( ) ( ) ( )2

' 2 32 2 2 2 1 21 1 2 1 2 2 21 1 2 1

1

1 1 12 1.5 cos cos 3 02 4 4e j j

ja f a a a Q aω γ σ σ γ γ α γ γ

=

+ − − − + + − =∑ (6.12.15d)

Where

( )1 1 1 1 2 2 1 2 11 , and 1.52

T Tγ β σ γ β σ σ= − + = − + −



The above equations are known as the reduced equations. For steady state, 0'2

'2

'1

'1 ==== γγ aa . So,

now we have a set of non-linear algebraic equations which is solved numerically to obtain the fixed point response of the system. The first-order solution of the system can be given by

( ) 11111 2/cos γτεσω −+= au (6.12.16a)

( ) [ ]221222 5.1cos γτσσεω −−+= au (6.12.16b)

Stability equations of steady-state response

By directly perturbing the reduced equations, one can study the stability of the non-trivial steady sate solution. But, as the reduced Eqs. (6.12.15a-d) have the coupled terms '

11γa and '22γa , the perturbed

equations will not contain the perturbations '1γ∆ or '

2γ∆ for trivial solutions and hence the stability of the trivial state cannot be studied by directly perturbing these equations. To circumvent this difficulty, normalization method is adopted by introducing the transformation

2,1,sin,cos === iaqap iiiiii γγ (6.12.17)

Into equations (6.12.15) to obtain the following normalized reduced equations or the Cartesian form of modulation equations:

( ) 2121111111'11 2

1212 qfqfpp +

−++ σωξω

( ) ( )∑=

=+−+−+2

1

2211121

21

21212 0

412

41

jjjje qpqqpppqqQ α (6.12.18a)

( ) 2121111111'11 2

1212 pfpfqq +

−++ σωξω

( ) ( )∑=

=+−+−+2

1

2211211

21

21212 0

412

41

jjjje qppqqpqppQ α (6.12.18b)

( ) ( ) 221212122'22 23

212 qqfpp σσωξω −+++

( ) ( )∑=

=−−−−2

1

2222

21

21121 0

413

41

jjjje qpqqpqQ α (6.12.18c)

( ) ( ) 221212122'22 23

212 ppfqq σσωξω −−−+



( ) ( )∑=

=++−+2

1

2222

21

21121 0

413

41

jjjje qppqppQ α (6.12.18d)

Now perturbing the above equations, one obtains

Tqpqp '2

,'2

'1

'1 ,,, ∆∆∆∆ [ ] T

c qpqpJ 2211 ,,, ∆∆∆∆= (6.12.19)

Where T is the transpose and [𝑗𝑐] is the Jacobian matrix whose eigenvalues will determine the stability and bifurcation of the system.

The stability boundary for the linear system (i.e. the trivial state) can be obtained from the eigen values of the matrix [𝐽𝑐] by letting 02121 ==== qqpp .

The first order solution of the system in terms of ( )2,1, =iqp ii can be given by

,sincos 11111 τωτω qpu += (6.12.20a)

τωτω 12122 3sin3cos qpu += , (6.12.20b)

where 111 21 εσωω += (6.12.21)

If the external frequency m n Ω=ω ±ω where nω is the nth natural frequency of the system one will

obtain combination resonance of sum ( )m nω ωΩ = + or difference ( )-m nω ωΩ = type for which one

may refer to the work of Dwivedy and Kar (1999). Also an exhaustive list of literature is given for the

interested reader.

Numerical Results and Discussion

Following Zavodney and Nayfeh [7] and keeping internal resonance in view, a metallic beam is considered with the following properties:

L=125.4 mm, I=0.04851 𝑚𝑚4, E=0.20936× 𝑁𝑚𝑚2 ,𝑍𝑟 = 1 𝑚𝑚, 𝑐 = 0.1 𝑁. 𝑠/𝑚𝑚2,

𝜌 = 0.03332 𝑔 𝑚𝑚 , 𝜇 = 3.68979, J=0.959, 𝛽 = 0.25

The roots of the characteristics equation are found numerically to be 𝑘1=1.80097, 𝑘2=3.2836 and the corresponding non-dimensional natural frequencies are 𝜔1=1 and 𝜔2=3.33179. The book keeping parameter 𝜀 and scaling factor λ are taken as 0.001 and 0.1, respectively. The coefficients of damping ( )nξ , excitation ( )mmf and non-linear terms ( )n

klmnklm

nklm γβα ,, are found to be of the same order. The



values of other required parameters expressed in the appendix (Kar and Dwivedy 1999) are calculated to be

𝛼𝑒11=2.54149, 𝛼𝑒12=-12.2027,𝛼𝑒21=-6.63699, 𝛼𝑒22=-195.55,𝑄12=14.62282, 𝑄21=7.84674,

∗11f =0.0655762, ∗

12f =0.0122118, ∗21f =0.04249, ∗

22f =0.1699298, ∗1ξ =0.0118963, ∗

2ξ =0.0045865

Figure 6.12.2 shows the trivial state instability regions for the system with principal parametric

resonance for different damping parameters. While the region bounded by the curves is unstable the

regions outside the curves are stable. Clearly due to the presence of internal resonance, in addition to

the main unstable region near φ =2, additional alternate zones of stable and unstable trivial branches

exists. With increase in damping and forcing amplitude these additional zones get merged with the

main unstable region. Here it may be noted that while with increase in damping the instability region

decreases, with increase in forcing amplitude, the instability region increases.

Fig. 6.12.2: A typical principal instability region



Fig. 6.12.3: Frequency response curve Γ 8, 0.001ν

A typical frequency response curve is shown in Figure 6.12.3 for both the first mode (lines without

bullet point) and the second mode (lines with bullet point). The stable branches are shown by solid lines

and the unstable branches are shown by dotted line. One may observe multi stable regions for a wide

range of frequency of the system. The nontrivial response amplitude of the first mode is observed to be

larger than the second mode. While supercritical and subcritical pitchfork bifurcations are observed in

the trivial state, both saddle node and Hopf bifurcations are observed in the nontrivial state. Due the

presence of Hopf bifurcation stable periodic response occurs in the trivial unstable region. With

decrease in forcing amplitude and damping parameter figure 6.12.4 shows the frequency response curve

for both the modes for Γ 5, 0.01ν . In addition to the other phenomenon described in the previous

figure, here one may clearly observe (in the insert) the additional alternate stable and unstable trivial

states near the main unstable region. While increasing the frequency one may observe jump up

phenomena and while decreasing the frequency one may observe the jump down phenomena in the

system.



Fig. 6.12.4: Frequency response curve for Γ 5, 0.01ν

Jump up Jump down

Jump up



Figure 6.12.5 shows the force response curve φ = 2.0, ν = 0.001.

Figure 6.12.6 shows the force response curve φ = 1.75, ν = 0.001.



Figure 6.12.7 Phase portraits: cascade of period doubling leading to chaos for φ = 2.13, Γ 8.0 , (a) ν = 8.5 (periodic), (b) ν = 8.4(2T periodic), ν = 8.3 (chaotic orbits).

Figure 6.12.8 Poincare’ showing period doubling route to chaos for φ = 2.13



Figure 6.12.5 shows the force response curve for the perfectly tuned system at φ = 2.0, ν = 0.001.

Herethe nontrivial fixed point becomes unstable with saddle nodde bifurcation points at Γ 0.9 and

1.6.And Hopf bifurcation points at Γ 5.75 . Here the trivial branch istotally unstable except at Γ 0

.Similarly Figure 6.12.6 shows the force response curve of the system at φ = 1.75, ν = 0.001. Though

the trivial response loses its stability at Γ 7.55 through supercritical Pitchfork bifurcation, and a

Hopf bifurcation is observed at Γ 7.95 , the system will fail through blue sky catastrophe if the

amplitude of excitation is increased beyond the

turning point at Γ 10.75 . Figure 6.12.7(a) shows the periodic response originating from the Hopf

bifurcation for φ = 2.13, Γ 8.0 ,ν = 8.5. With decrease in the damping parameter ν to 8.42 one may

observe a response with double period (Figure 6.12.7(b)). This period doubling phenomena continues

with further decrease in damping parameter and finally a chaotic response (Figure 6.12.7(c)) is

observed. Figure 6.12.8 shows the Poincare’ section depicting cascade of period-doubling leading to

chaos.

Reference for further study

1. K.L. Hadoo, V. Sundarajan, Parametric instability of a cantilevered column with end mass,

Journal of Sound and Vibration 18, 45-53, 1971

2. E.G. Tezak, D.T. Mook, A.H. Nayfeh, Nonlinear analysis of the lateral response of coloums to

periodic loads, ASME Journal of Machine Design, 100, 651-655, 1978.

3. C.W.S. To, Vibration of a cantilever beam with a base excitation and tip mass, Journal of Sound

and Vibration 83, 445-460, 1982.

4. A.H. Nayfeh, The response of multi degree freedom systems with quardtic non-linearities to a

harmonic parametric resonance, Journal of Sound and Vibration, 90, 237-244, 1983.

5. A.H. Nayfeh, A.E.S. Jebrill, The resonance of two degree of freedom systems with quardtic

non-linearities to parametrtic excitation, Journal of Sound and Vibration 88, 547-557, 1983.

6. L. D. Zavodney, A theoretical and experimental investigation of parametrically excited non-

linear mechanical system. PhD Dissertation, Dept. of. Engineering Science and Mechanics,

Virginia Polytechnic and State University, Blacksburg, December 1987.



7. L. D. Zavodney and A. H. Nayfeh, The nonlinear response of a slender beam carrying a lumped

mass to a principal parametric excitation: theory and experiment, International Journal of Non-

linear Mechanics, 24, 105-125, 1989.

8. A.H. Nayfeh, P.F. Pai, Nonlinear non-planar parametric responses of an in-extensional beam,

International Journal of Non-linear Mechanics, 24, 139-158, 1989.

9. A.H. Nayfeh, B. Balachandram, Modal interactions in dynamical and structrual systems, ASME

Applied Mechanics Review, 42, s175-s201, 1989.

10. H.P. Lee, Stability of a cantilever beam with tip mass subjected to axial sinusoidal excitations,

Journal of Sound and Vibration 183, 91-98, 1995.

11. B. Banarjee, A.K. Bajaj, P. Davies, Resonant dynamics of an autoparametric systems: a study

using higher order avearging, International Journal Mechanical, 31, 21-39, 1996.

12. E.H.K. Fung, Z.X. Shi, Vibration frequencies of a constrained flexible arm carrying an end

mass, Journal of Sound and Vibration 203, 373-387, 1997.

13. M.N. Hamdan, M.H.F. Dado, Large amplitude free vibrations of a uniform cantilever beam

carrrying an intermediate lumped mass and rotary inertia, Journal of Sound and Vibration 206,

151-168, 1997.

14. H. Erol, M. Gürgöze, On The Effect of an Attached Spring-mass System On The Frequency

Spectrum Of Longitudinally Vibrating Elastic Rods Carrying A Tip Mass, Journal of Sound and

Vibration, 227, (2) 471-477, 1999.

15. S. Inceo Ǧ, M. Gürgöze, Bending Vibrations Of Beams Coupled By Several Double Spring-

mass Systems, Journal of Sound and Vibration, 243 (2), 370-379, 2001.

16. H Erol, M Gürgöze, Longitudinal vibrations of a double-rod system coupled by springs and

dampers, Journal of Sound and Vibration, 276 (1–2), 419-430, 2004.

17. M. Gürgöze, H. Erol , On laterally vibrating beams carrying tip masses, coupled by several

double spring–mass systems, Journal of Sound and Vibration, 269 (1–2), 431-438, 2004.

18. M. Gürgöze On the representation of a cantilevered beam carrying a tip mass by an equivalent

spring–mass system Journal of Sound and Vibration, 282, (1–2), 538-542, 2005.

19. S. K. Dwivedy and R. C. Kar, Dynamics of a slender beam with an attached mass under

combination parametric and internal resonances, Part II: Periodic and Chaotic response. Journal

of Sound and Vibration, 222, no 2, 281-305, 1999.



20. S. K. Dwivedy and R. C. Kar, Dynamics of a slender beam with an attached mass under

combination parametric and internal resonances, Part I: steady state response. Journal of Sound

and Vibration, 221, no 5, 823-848, 1999

21. R. C. Kar and S. K. Dwivedy, Non-linear dynamics of a slender beam carrying a lumped mass

with principal parametric and internal resonances. International Journal of Nonlinear

Mechanics, 34, (3), 515-529, 1999.

22. S. K. Dwivedy and R. C. Kar, Simultaneous combination, principal parametric and internal

resonances in a slender beam with a lumped mass: three mode interactions. Journal of Sound

and Vibration Vol. 242, no 1 pp 27-46, 2001.

23. S. K. Dwivedy and R. C. Kar, Non-linear dynamics of a slender beam carrying a lumped mass

under principal parametric resonances with three-mode interactions. International Journal of

Nonlinear Mechanics, Vol. 36, no. 6, pp. 927-945, 2001.

24. S. K. Dwivedy and R. C. Kar, Nonlinear response of a parametrically excited system using

higher order multiple scales. Nonlinear Dynamics, 20, 115-130, 1999.

25. S. K. Dwivedy and R. C. Kar, Nonlinear dynamics of a cantilever beam carrying an attached

mass with 1:3:9 internal resonances. Nonlinear Dynamics, 31(1), 49-72, 2003.

26. S. K. Dwivedy and R. C. Kar, Simultaneous combination and 1:3:5 internal resonances in a

parametrically excited beam-mass system. International Journal of Nonlinear Mechanics, 38,

(4), 585-596, 2003.

27. S. K. Dwivedy and R. C. Kar, Nonlinear response of a parametrically excited slender beam

carrying a lumped mass with 1:3:9 internal resonance, Advances in Vibration Engineering. 2,

(3), 1-9, 2003.

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